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Calculus 3 : Relative Minimums and Maximums

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Example Question #121 : Applications Of Partial Derivatives

Find and classify all the critical points for $\displaystyle f(x,y)=5x^2y+3y^2+2x^2-4y^3+1$.

Possible Answers:

(0,0): $\displaystyle (0,0):$ Relative Maximum

$(0,\frac{1}{2}):$ $\displaystyle (0,\frac{1}{2}):$ Relative Minimum

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

(0,0): $\displaystyle (0,0):$ Saddle Point

$(0,\frac{1}{2}):$ $\displaystyle (0,\frac{1}{2}):$ Saddle Point

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

(0,0): $\displaystyle (0,0):$ Relative Minimum

$(0,\frac{1}{2}):$ $\displaystyle (0,\frac{1}{2}):$ Relative Minimum

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

(0,0): $\displaystyle (0,0):$ Relative Minimum

$(0,\frac{1}{2}):$ $\displaystyle (0,\frac{1}{2}):$ Saddle Point

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

(0,0): $\displaystyle (0,0):$ Relative Minimum

$(0,\frac{1}{2}):$ $\displaystyle (0,\frac{1}{2}):$ Saddle Point

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Relative Maximum

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Relative Minimum

Correct answer:

(0,0): $\displaystyle (0,0):$ Relative Minimum

$(0,\frac{1}{2}):$ $\displaystyle (0,\frac{1}{2}):$ Saddle Point

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ Saddle Point

Explanation:

First thing we need to do is take partial derivatives.

$\displaystyle f_x(x,y)=10xy+4x$

$f_{xx}(x,y)=10y+4$ $\displaystyle f_{xx}(x,y)=10y+4$

$\displaystyle f_y(x,y)=5x^2+6y-12y^2$

$f_{yy}(x,y)=6-24y$ $\displaystyle f_{yy}(x,y)=6-24y$

$f_{xy}(x,y)=10x$ $\displaystyle f_{xy}(x,y)=10x$

Now we want to find critical points, we do this by setting the partial derivative in respect to x equal to zero.

$10xy+4x=0 \rightarrow 2x(5y+2)=0\rightarrow x=0, y=-\frac{2}{5}$ $\displaystyle 10xy+4x=0 \rightarrow 2x(5y+2)=0\rightarrow x=0, y=-\frac{2}{5}$

Now we want to plug in these values into the partial derivative in respect to y and set it equal to zero.

x=0: $\displaystyle x=0:$

$5(0)^2+6y-12y^2=0\rightarrow 6y(1-2y)=0\rightarrow y=0, y=\frac{1}{2}$ $\displaystyle 5(0)^2+6y-12y^2=0\rightarrow 6y(1-2y)=0\rightarrow y=0, y=\frac{1}{2}$

$y=-\frac{2}{5}:$ $\displaystyle y=-\frac{2}{5}:$

$5x^2+6(-\frac{2}{5})-12(\frac{2}{5})^2=0\rightarrow x^2=\frac{108}{125}\rightarrow x=\pm\sqrt{\frac{108}{125}}$ $\displaystyle 5x^2+6(-\frac{2}{5})-12(\frac{2}{5})^2=0\rightarrow x^2=\frac{108}{125}\rightarrow x=\pm\sqrt{\frac{108}{125}}$

Lets summarize the critical points:

If x=0: $\displaystyle x=0:$

$(0,0), (0, \frac{1}{2})$ $\displaystyle (0,0), (0, \frac{1}{2})$

If $y=-\frac{2}{5}:$ $\displaystyle y=-\frac{2}{5}:$

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big), \Big(-\sqrt{\frac{108}{125}},- \frac{2}{5}\Big)$ $\displaystyle \Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big), \Big(-\sqrt{\frac{108}{125}},- \frac{2}{5}\Big)$

Now we need to classify these points, we do this by creating a general formula $\displaystyle D$.

$D=f_{xx}(a,b)f_{yy}(a,b)-(f_{xy}(a,b))^2$ $\displaystyle D=f_{xx}(a,b)f_{yy}(a,b)-(f_{xy}(a,b))^2$, where a,b $\displaystyle a,b$, is a critical point.

If D>0 $\displaystyle D>0$ and $f_{xx}(a,b)>0$ $\displaystyle f_{xx}(a,b)>0$, then there is a relative minimum at (a,b) $\displaystyle (a,b)$

If D>0 $\displaystyle D>0$ and $f_{xx}(a,b)<0$ $\displaystyle f_{xx}(a,b)< 0$, then there is a relative maximum at (a,b) $\displaystyle (a,b)$

If D<0 $\displaystyle D< 0$, there is a saddle point at (a,b) $\displaystyle (a,b)$

If D=0 $\displaystyle D=0$ then the point (a,b) $\displaystyle (a,b)$ may be a relative minimum, relative maximum or a saddle point.

$\displaystyle D=(10y+4)(6-24y)-(10x)^2$

Now we plug in the critical values into $\displaystyle D$.

(0,0): $\displaystyle (0,0):$

$\displaystyle D=(10(0)+4)(6-24(0))-(10(0))^2=(4)(6)=24$

$f_{xx}(0,0)=10(0)+4=4$ $\displaystyle f_{xx}(0,0)=10(0)+4=4$

Since D>0 $\displaystyle D>0$ and $f_{xx}(a,b)>0$ $\displaystyle f_{xx}(a,b)>0$, (0,0) $\displaystyle (0,0)$ is a relative minimum.

$(0,\frac{1}{2}):$ $\displaystyle (0,\frac{1}{2}):$

$D=(10(\frac{1}{2})+4)(6-24(\frac{1}{2}))-(10(0))^2=(5+4)(6-12)=-54$ $\displaystyle D=(10(\frac{1}{2})+4)(6-24(\frac{1}{2}))-(10(0))^2=(5+4)(6-12)=-54$

Since D<0 $\displaystyle D< 0$, $(0,\frac{1}{2})$ $\displaystyle (0,\frac{1}{2})$ is a saddle point.

$\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$

$D=(10(-\frac{2}{5})+4)(6-24(-\frac{2}{5}))-(10(\sqrt{\frac{108}{125}}))^2=(0)-\frac{432}{5}=-\frac{432}{5}$ $\displaystyle D=(10(-\frac{2}{5})+4)(6-24(-\frac{2}{5}))-(10(\sqrt{\frac{108}{125}}))^2=(0)-\frac{432}{5}=-\frac{432}{5}$

Since D<0 $\displaystyle D< 0$, $\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big)$ $\displaystyle \Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big)$ is a saddle point

$\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$ $\displaystyle \Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big):$

$D=(10(-\frac{2}{5})+4)(6-24(-\frac{2}{5}))-(10(-\sqrt{\frac{108}{125}}))^2=(0)-\frac{432}{5}=-\frac{432}{5}$ $\displaystyle D=(10(-\frac{2}{5})+4)(6-24(-\frac{2}{5}))-(10(-\sqrt{\frac{108}{125}}))^2=(0)-\frac{432}{5}=-\frac{432}{5}$

Since D<0 $\displaystyle D< 0$, $\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big)$ $\displaystyle \Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big)$ is a saddle point

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Example Question #2 : Relative Minimums And Maximums

Find the relative maxima and minima of $\displaystyle f(x,y)=5x^2+5y^2+4xy-3$.

Possible Answers:

(5,5) $\displaystyle (5,5)$ is a relative maximum.

(0,0) $\displaystyle (0,0)$ is a relative minimum.

(0,0) $\displaystyle (0,0)$ is a relative maximum.

(5,5) $\displaystyle (5,5)$ is a relative minimum.

Correct answer:

(0,0) $\displaystyle (0,0)$ is a relative minimum.

Explanation:

To find the relative maxima and minima, we must find all the first order and second order partial derivatives. We will use the first order partial derivative to find the critical points, then use the equation

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)<0$ $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If D(x,y)<0 $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=10x+4y$ $\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=10x+4y$

$\frac{\partial f }{\partial y}=f_y(x,y)=10y+4x$ $\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=10y+4x$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=10$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=10$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=10$ $\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=10$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=4$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=4$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$\displaystyle f_x(x,y)=10x+4y=0$

$\displaystyle f_y(x,y)=10y+4x=0$

$10x+4y=0\Rightarrow 10x=-4y\Rightarrow x=-4y/10$ $\displaystyle 10x+4y=0\Rightarrow 10x=-4y\Rightarrow x=-4y/10$

$10y+4x=0\Rightarrow 10y+4(-4y/10)=0 \Rightarrow 10y-16y/10=0$ $\displaystyle 10y+4x=0\Rightarrow 10y+4(-4y/10)=0 \Rightarrow 10y-16y/10=0$

$\displaystyle 84y/10=0$

y=0 $\displaystyle y=0$

$\displaystyle x=-4y/10=-4(0)/10=0$

There is only one critical point and it is at (0,0) $\displaystyle (0,0)$. We need to determine if this critical point is a maximum or minimum using D(x,y) $\displaystyle D(x,y)$ and $f_{xx}(x,y)$ $\displaystyle f_{xx}(x,y)$.

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=10(10)-(4)^2=100-16=84$

$f_{xx}(x,y)=10$ $\displaystyle f_{xx}(x,y)=10$

Since $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, (0,0) $\displaystyle (0,0)$ is a relative minimum.

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Example Question #3 : Relative Minimums And Maximums

Find the relative maxima and minima of $\displaystyle f(x,y)=x^3+4y^3-2xy+17$.

Possible Answers:

(0,0) $\displaystyle (0,0)$ and $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ $\displaystyle \left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ are relative maxima

(0,0) $\displaystyle (0,0)$ and $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ $\displaystyle \left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ are relative minima

(0,0) $\displaystyle (0,0)$ is a saddle point, $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ $\displaystyle \left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ is a relative minimum

(0,0) $\displaystyle (0,0)$ is a relative maximum, $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ $\displaystyle \left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ is a relative minimum

Correct answer:

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)<0$ $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If D(x,y)<0 $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=3x^2-2y$ $\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=3x^2-2y$

$\frac{\partial f }{\partial y}=f_y(x,y)=12y^2-2x$ $\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=12y^2-2x$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=6x$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=6x$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=24y$ $\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=24y$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=-2$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=-2$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$f_x(x,y)=3x^2-2y=0\Rightarrow 2y=3x^2\Rightarrow y=3x^2/2$ $\displaystyle f_x(x,y)=3x^2-2y=0\Rightarrow 2y=3x^2\Rightarrow y=3x^2/2$

$\displaystyle f_y(x,y)=12y^2-2x=0$

$f_y(x,y)=12(3x^2/2)^2-2x=0\Rightarrow 27x^4-2x=0$ $\displaystyle f_y(x,y)=12(3x^2/2)^2-2x=0\Rightarrow 27x^4-2x=0$

$\displaystyle x(27x^3-2)=0$

There are two possible values of $\displaystyle x$, x=0 $\displaystyle x=0$ and $x=\frac{\sqrt[3]{2}}{3}$ $\displaystyle x=\frac{\sqrt[3]{2}}{3}$.

We find the corresponding values of $\displaystyle y$ using $\displaystyle y=3x^2/2$ (found by rearranging the first derivative)

$\displaystyle y(0)=3(0)^2/2=0$

$y=\frac{3\left ( \frac{\sqrt[3]{2}}{3} \right )^2}{2} =\frac{3\left ( \frac{\sqrt[3]{4}}{9} \right )}{2} =\frac{\sqrt[3]{4}}{6}$ $\displaystyle y=\frac{3\left ( \frac{\sqrt[3]{2}}{3} \right )^2}{2} =\frac{3\left ( \frac{\sqrt[3]{4}}{9} \right )}{2} =\frac{\sqrt[3]{4}}{6}$

There are critical points at (0,0) $\displaystyle (0,0)$ and $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ $\displaystyle \left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$. We need to determine if the critical points are maximums or minimums using D(x,y) $\displaystyle D(x,y)$ and $f_{xx}(x,y)$ $\displaystyle f_{xx}(x,y)$.

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$D(x,y)=(6x)(24y)-\left ( -2 \right )^2=144xy-4$ $\displaystyle D(x,y)=(6x)(24y)-\left ( -2 \right )^2=144xy-4$

At (0,0) $\displaystyle (0,0)$,

D(x,y)=144(0)(0)-4=-4<0 $\displaystyle D(x,y)=144(0)(0)-4=-4< 0$

Since D(x,y)<0 $\displaystyle D(x,y)< 0$, (0,0) $\displaystyle (0,0)$ is a saddle point.

At $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ $\displaystyle \left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$,

$D(x,y)=144\left ( \frac{\sqrt[3]{2}}{3} \right )\left ( \frac{\sqrt[3]{4}}{6} \right )-4=144\left ( \frac{\sqrt[3]{8}}{18} \right )-4$ $\displaystyle D(x,y)=144\left ( \frac{\sqrt[3]{2}}{3} \right )\left ( \frac{\sqrt[3]{4}}{6} \right )-4=144\left ( \frac{\sqrt[3]{8}}{18} \right )-4$

$=144\left ( \frac{2}{18} \right )-4=12>0$ $\displaystyle =144\left ( \frac{2}{18} \right )-4=12>0$

$f_{xx}(x,y)=6\left ( \frac{\sqrt[3]{2}}{3} \right )=2\sqrt[3]{2}>0$ $\displaystyle f_{xx}(x,y)=6\left ( \frac{\sqrt[3]{2}}{3} \right )=2\sqrt[3]{2}>0$

Since $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, $\left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ $\displaystyle \left (\frac{\sqrt[3]{2}}{3},\frac{\sqrt[3]{4}}{6} \right )$ is a relative minimum.

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Example Question #2 : Relative Minimums And Maximums

Find the relative maxima and minima of $\displaystyle f(x,y)=x^2y+4y^3+xy+1$.

Possible Answers:

(0,0) $\displaystyle (0,0)$, (-1,0) $\displaystyle (-1,0)$ and $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ $\displaystyle \left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ are relative maxima.

(0,0) $\displaystyle (0,0)$ and (-1,0) $\displaystyle (-1,0)$ are relative minima, $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ $\displaystyle \left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ is a relative maximum.

Correct answer:

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)<0$ $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If D(x,y)<0 $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=2xy+y$ $\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=2xy+y$

$\frac{\partial f }{\partial y}=f_y(x,y)=x^2+12y^2+x$ $\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=x^2+12y^2+x$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2y$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2y$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=2x+1$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=2x+1$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$f_y(x,y)=x^2-12y^2+x=0\Rightarrow 12y^2=-x^2-x$ $\displaystyle f_y(x,y)=x^2-12y^2+x=0\Rightarrow 12y^2=-x^2-x$

$y=\sqrt{\frac{-x(x+1)}{12}}$ $\displaystyle y=\sqrt{\frac{-x(x+1)}{12}}$

$f_x(x,y)=2xy+y=0\Rightarrow y(2x+1)=0$ $\displaystyle f_x(x,y)=2xy+y=0\Rightarrow y(2x+1)=0$

$\sqrt{\frac{-x(x+1)}{12}}(2x+1)=0$ $\displaystyle \sqrt{\frac{-x(x+1)}{12}}(2x+1)=0$

Squaring both sides of the equation gives us

$\frac{-x(x+1)}{12}(2x+1)^2=0^2$ $\displaystyle \frac{-x(x+1)}{12}(2x+1)^2=0^2$

Multiplying both sides of the equation by $\displaystyle 12$ gives us

$\displaystyle -x(x+1)(2x+1)^2=0$

There are three possible values of $\displaystyle x$; x=0 $\displaystyle x=0$, x=-1 $\displaystyle x=-1$ and x=-1/2 $\displaystyle x=-1/2$.

We find the corresponding values of $\displaystyle y$ using $y=\sqrt{\frac{-x(x+1)}{12}}$ $\displaystyle y=\sqrt{\frac{-x(x+1)}{12}}$ (found by rearranging the first derivative)

$y(0)=\sqrt{\frac{-(0)(0+1)}{12}}=\sqrt{\frac{0}{12}}=0$ $\displaystyle y(0)=\sqrt{\frac{-(0)(0+1)}{12}}=\sqrt{\frac{0}{12}}=0$

$y(-1)=\sqrt{\frac{-(-1)(-1+1)}{12}}=\sqrt{\frac{-(-1)(0)}{12}}=0$ $\displaystyle y(-1)=\sqrt{\frac{-(-1)(-1+1)}{12}}=\sqrt{\frac{-(-1)(0)}{12}}=0$

$y(-1/2)=\sqrt{\frac{-(-1/2)(-1/2+1)}{12}}=\sqrt{\frac{-(-1/2)(1/2)}{12}}$ $\displaystyle y(-1/2)=\sqrt{\frac{-(-1/2)(-1/2+1)}{12}}=\sqrt{\frac{-(-1/2)(1/2)}{12}}$

$=\sqrt{\frac{(1/4)}{12}}=\sqrt{\frac{1}{48}}$ $\displaystyle =\sqrt{\frac{(1/4)}{12}}=\sqrt{\frac{1}{48}}$

There are critical points at (0,0) $\displaystyle (0,0)$, (-1,0) $\displaystyle (-1,0)$ and $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ $\displaystyle \left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$. We need to determine if the critical points are maximums or minimums using D(x,y) $\displaystyle D(x,y)$ and $f_{xx}(x,y)$ $\displaystyle f_{xx}(x,y)$.

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=(2y)(24y)-(2x+1)^2=48y^2-(4x^2+4x+1)$

$\displaystyle D(x,y)=48y^2-4x^2-4x-1$

At (0,0) $\displaystyle (0,0)$,

$\displaystyle D(x,y)=48(0)^2-4(0)^2-4(0)-1=-1< 0$

Since D(x,y)<0 $\displaystyle D(x,y)< 0$, (0,0) $\displaystyle (0,0)$ is a saddle point.

At (-1,0) $\displaystyle (-1,0)$,

$\displaystyle D(x,y)=48(0)^2-4(-1)^2-4(-1)-1=0-4-4-1$

D(x,y)=-9<0 $\displaystyle D(x,y)=-9< 0$

Since D(x,y)<0 $\displaystyle D(x,y)< 0$, (-1,0) $\displaystyle (-1,0)$ is a saddle point.

At $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ $\displaystyle \left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$,

$D(x,y)=48\left ( \sqrt{\frac{1}{48}} \right )^2-4(1/2)^2-4(1/2)-1$ $\displaystyle D(x,y)=48\left ( \sqrt{\frac{1}{48}} \right )^2-4(1/2)^2-4(1/2)-1$

$D(x,y)=48\left ( \frac{1}{48} \right )-4(1/4)-4(1/2)-1$ $\displaystyle D(x,y)=48\left ( \frac{1}{48} \right )-4(1/4)-4(1/2)-1$

D(x,y)=1-1-2-1=-3<0 $\displaystyle D(x,y)=1-1-2-1=-3< 0$

Since D(x,y)<0 $\displaystyle D(x,y)< 0$, $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ $\displaystyle \left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ is a saddle point.

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Example Question #5 : Relative Minimums And Maximums

Find the relative maxima and minima of $\displaystyle f(x,y)=(x+4)^2-2(y-1)^2+3x^2y+17$.

Possible Answers:

(0,0) $\displaystyle (0,0)$ and $\displaystyle (1.429,-0.757)$ are relative minima.

$\displaystyle (1.429,-0.757)$ is a relative minimum.

$\displaystyle (1.429,-0.757)$ is a relative maximum.

$\displaystyle (1.429,-0.757)$ is a saddle point.

Correct answer:

$\displaystyle (1.429,-0.757)$ is a saddle point.

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)<0$ $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If D(x,y)<0 $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=2(x+4)+6xy=2x+8+6xy$ $\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=2(x+4)+6xy=2x+8+6xy$

$\frac{\partial f }{\partial y}=f_y(x,y)=-4(y-1)+3x^2=3x^2-4y+4$ $\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=-4(y-1)+3x^2=3x^2-4y+4$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2+6y$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2+6y$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=-4$ $\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=-4$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=6x$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=6x$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$f_x(x,y)=2x+8+6xy=0\Rightarrow 2x+6xy=-8$ $\displaystyle f_x(x,y)=2x+8+6xy=0\Rightarrow 2x+6xy=-8$

$2x+6xy=-8\Rightarrow x+3xy=-4\Rightarrow x(1+3y)=-4$ $\displaystyle 2x+6xy=-8\Rightarrow x+3xy=-4\Rightarrow x(1+3y)=-4$

$x=\frac{-4}{1+3y}$ $\displaystyle x=\frac{-4}{1+3y}$

$f_y(x,y)=3x^2-4y+4=0\Rightarrow 3\left ( \frac{-4}{1+3y} \right )^2-4y+4=0$ $\displaystyle f_y(x,y)=3x^2-4y+4=0\Rightarrow 3\left ( \frac{-4}{1+3y} \right )^2-4y+4=0$

$3\left ( \frac{16}{(1+3y)^2} \right )-4y+4=0$ $\displaystyle 3\left ( \frac{16}{(1+3y)^2} \right )-4y+4=0$

$\frac{48}{(1+3y)^2}=4y-4$ $\displaystyle \frac{48}{(1+3y)^2}=4y-4$

$\displaystyle 48=(4y-4)(1+3y)^2$

$\displaystyle (4y-4)(1+3y)^2-48=0$

$\displaystyle (4y-4)(1+6y+9y^2)-48=0$

$\displaystyle 4y+24y^2+36y^3-4-24y-36y^2-48=0$

$\displaystyle 36y^3-12y^2-20y-52=0$

There is only one real value of $\displaystyle y$; $y\approx1.429$ $\displaystyle y\approx1.429$

We find the corresponding value of $\displaystyle x$ using $x=\frac{-4}{1+3y}$ $\displaystyle x=\frac{-4}{1+3y}$ (found by rearranging the first derivative)

$x(1.429)=\frac{-4}{1+3(1.429)}\approx -0.757$ $\displaystyle x(1.429)=\frac{-4}{1+3(1.429)}\approx -0.757$

There is a critical point at $\displaystyle (1.429,-0.757)$. We need to determine if the critical point is a maximum or minimum using D(x,y) $\displaystyle D(x,y)$ and $f_{xx}(x,y)$ $\displaystyle f_{xx}(x,y)$.

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=(2+6y)(-4)-(6x)^2$

$\displaystyle D(x,y)=-8-24y-36x^2$

At $\displaystyle (1.429,-0.757)$,

$\displaystyle D(x,y)=-8-24(-0.757)-36(1.429)^2=-63.35< 0$

Since D(x,y)<0 $\displaystyle D(x,y)< 0$, $\displaystyle (1.429,-0.757)$ is a saddle point.

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Example Question #6 : Relative Minimums And Maximums

Find the relative maxima and minima of $\displaystyle f(x,y)=y^4+2x^3-2x^2y-7xy+9$.

Possible Answers:

(0,0) $\displaystyle (0,0)$ is a relative minima, $\displaystyle (-1.6,-1.786)$ is a relative maxima

(0,0) $\displaystyle (0,0)$ and $\displaystyle (-1.6,-1.786)$ are relative maxima

(0,0) $\displaystyle (0,0)$ is a relative maxima, $\displaystyle (-1.6,-1.786)$ is a relative minima

(0,0) $\displaystyle (0,0)$ and $\displaystyle (-1.6,-1.786)$ are saddle points

Correct answer:

(0,0) $\displaystyle (0,0)$ and $\displaystyle (-1.6,-1.786)$ are saddle points

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)<0$ $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If D(x,y)<0 $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=6x^2-4xy-7y$ $\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=6x^2-4xy-7y$

$\frac{\partial f }{\partial y}=f_y(x,y)=4y^3-2x^2-7x$ $\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=4y^3-2x^2-7x$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=12x-4y$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=12x-4y$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=12y^2$ $\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=12y^2$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=-4x-7$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=-4x-7$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$f_x(x,y)=6x^2-4xy-7y=0\Rightarrow 6x^2-y(4x-7)=0$ $\displaystyle f_x(x,y)=6x^2-4xy-7y=0\Rightarrow 6x^2-y(4x-7)=0$

$6x^2=y(4x-7)\Rightarrow y=\frac{6x^2}{4x-7}$ $\displaystyle 6x^2=y(4x-7)\Rightarrow y=\frac{6x^2}{4x-7}$

$f_y(x,y)=4y^3-2x^2-7x=0\Rightarrow 4\left ( \frac{6x^2}{4x-7} \right )^3-2x^2-7x=0$ $\displaystyle f_y(x,y)=4y^3-2x^2-7x=0\Rightarrow 4\left ( \frac{6x^2}{4x-7} \right )^3-2x^2-7x=0$

$\displaystyle 4(6x^2)^3-2x^2(4x-7)^3-7x(4x-7)^3=0(4x-7)^3$

$\displaystyle 4(6x^2)^3-(2x^2+7x)(4x-7)^3=0$

$\displaystyle 864x^6-128x^5+224x^4+1176x^3-3430x^2+2401x=0$

The real values of $\displaystyle x$ are x=0 $\displaystyle x=0$ and x=-1.6 $\displaystyle x=-1.6$

We find the corresponding value of $\displaystyle y$ using $y=\frac{6x^2}{4x-7}$ $\displaystyle y=\frac{6x^2}{4x-7}$ (found by rearranging the first derivative)

$y(0)=\frac{6(0)^2}{4(0)-7}=0$ $\displaystyle y(0)=\frac{6(0)^2}{4(0)-7}=0$

$y(-1.6)=\frac{6(-1.6)^2}{4(-1.6)-7}=\frac{6*2.56}{-1.6-7}\approx -1.786$ $\displaystyle y(-1.6)=\frac{6(-1.6)^2}{4(-1.6)-7}=\frac{6*2.56}{-1.6-7}\approx -1.786$

There are critical points at (0,0) $\displaystyle (0,0)$ and $\displaystyle (-1.6,-1.786)$. We need to determine if the critical points are maxima or minima using D(x,y) $\displaystyle D(x,y)$ and $f_{xx}(x,y)$ $\displaystyle f_{xx}(x,y)$.

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=(12x-4y)(12y^2)-(-4x-7)^2$

At (0,0) $\displaystyle (0,0)$,

$\displaystyle D(x,y)=(12(0)-4(0))(12(0)^2)-(-4(0)-7)^2=0-(-7)^2=-49< 0$

Since D(x,y)<0 $\displaystyle D(x,y)< 0$, (0,0) $\displaystyle (0,0)$ is a saddle point.

At $\displaystyle (-1.6,-1.786)$,

$\displaystyle D(x,y)=(12(-1.6)-4(-1.786))(12(-1.786)^2)-(-4(-1.6)-7)^2$

$\displaystyle D(x,y)=(-19.2-7.144)(38.278)-(6.4-7)^2=-1008.7< 0$

Since D(x,y)<0 $\displaystyle D(x,y)< 0$, $\displaystyle (-1.6,-1.786)$ is a saddle point.

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Example Question #3 : Relative Minimums And Maximums

Find the relative maxima and minima of $\displaystyle f(x,y)=x^4+y^4+3xy$.

Possible Answers:

(0,0) $\displaystyle (0,0)$ is a saddle point, $(-\sqrt3/2,\sqrt3/2)$ $\displaystyle (-\sqrt3/2,\sqrt3/2)$ and $(\sqrt3/2,-\sqrt3/2)$ $\displaystyle (\sqrt3/2,-\sqrt3/2)$ are relative minima.

(0,0) $\displaystyle (0,0)$ is a saddle point, $(-\sqrt3/2,\sqrt3/2)$ $\displaystyle (-\sqrt3/2,\sqrt3/2)$ and $(\sqrt3/2,-\sqrt3/2)$ $\displaystyle (\sqrt3/2,-\sqrt3/2)$ are saddle points.

(0,0) $\displaystyle (0,0)$ is a relative minima, $(-\sqrt3/2,\sqrt3/2)$ $\displaystyle (-\sqrt3/2,\sqrt3/2)$ and $(\sqrt3/2,-\sqrt3/2)$ $\displaystyle (\sqrt3/2,-\sqrt3/2)$ are relative maxima.

(0,0) $\displaystyle (0,0)$, $(-\sqrt3/2,\sqrt3/2)$ $\displaystyle (-\sqrt3/2,\sqrt3/2)$ and $(\sqrt3/2,-\sqrt3/2)$ $\displaystyle (\sqrt3/2,-\sqrt3/2)$ are relative maxima.

Correct answer:

(0,0) $\displaystyle (0,0)$ is a saddle point, $(-\sqrt3/2,\sqrt3/2)$ $\displaystyle (-\sqrt3/2,\sqrt3/2)$ and $(\sqrt3/2,-\sqrt3/2)$ $\displaystyle (\sqrt3/2,-\sqrt3/2)$ are relative minima.

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)<0$ $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If D(x,y)<0 $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=4x^3+3y$ $\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=4x^3+3y$

$\frac{\partial f }{\partial y}=f_y(x,y)=4y^3+3x$ $\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=4y^3+3x$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=12x^2$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=12x^2$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=3$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=3$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$f_x(x,y)=4x^3+3y=0\Rightarrow 4x^3=-3y \Rightarrow y=\frac{-4x^3}{3}$ $\displaystyle f_x(x,y)=4x^3+3y=0\Rightarrow 4x^3=-3y \Rightarrow y=\frac{-4x^3}{3}$

$f_y(x,y)=4y^3+3x=0\Rightarrow 4\left ( \frac{-4x^3}{3} \right )^3+3x=0$ $\displaystyle f_y(x,y)=4y^3+3x=0\Rightarrow 4\left ( \frac{-4x^3}{3} \right )^3+3x=0$

$4\left ( \frac{-64x^9}{27} \right )+3x=0\Rightarrow \frac{-256}{27}x^9+3x=0$ $\displaystyle 4\left ( \frac{-64x^9}{27} \right )+3x=0\Rightarrow \frac{-256}{27}x^9+3x=0$

$x\left (\frac{-256}{27}x^8+3 \right )=0$ $\displaystyle x\left (\frac{-256}{27}x^8+3 \right )=0$

Setting each factor in the expression equal to $\displaystyle 0$ gives us

x=0 $\displaystyle x=0$ and $\frac{-256}{27}x^8+3=0\Rightarrow\frac{-256}{27}x^8=-3$ $\displaystyle \frac{-256}{27}x^8+3=0\Rightarrow\frac{-256}{27}x^8=-3$

$x=\sqrt[8]{-3*\frac{27}{-257}}=\pm \sqrt3/2$ $\displaystyle x=\sqrt[8]{-3*\frac{27}{-257}}=\pm \sqrt3/2$

The real values of $\displaystyle x$ are x=0 $\displaystyle x=0$, $x=-\sqrt3/2$ $\displaystyle x=-\sqrt3/2$ and $x=\sqrt3/2$ $\displaystyle x=\sqrt3/2$

We find the corresponding value of $\displaystyle y$ using $y=\frac{-4x^3}{3}$ $\displaystyle y=\frac{-4x^3}{3}$ (found by rearranging the first derivative)

$y(0)=\frac{-4(0)^3}{3}=0$ $\displaystyle y(0)=\frac{-4(0)^3}{3}=0$

$y(-\sqrt3/2)=\frac{-4(-\sqrt3/2)^3}{3}=\sqrt3/2$ $\displaystyle y(-\sqrt3/2)=\frac{-4(-\sqrt3/2)^3}{3}=\sqrt3/2$

$y(\sqrt3/2)=\frac{-4(-\sqrt3/2)^3}{3}=-\sqrt3/2$ $\displaystyle y(\sqrt3/2)=\frac{-4(-\sqrt3/2)^3}{3}=-\sqrt3/2$

There are critical points at (0,0) $\displaystyle (0,0)$, $(-\sqrt3/2,\sqrt3/2)$ $\displaystyle (-\sqrt3/2,\sqrt3/2)$ and $(\sqrt3/2,-\sqrt3/2)$ $\displaystyle (\sqrt3/2,-\sqrt3/2)$. We need to determine if the critical points are maxima or minima using D(x,y) $\displaystyle D(x,y)$ and $f_{xx}(x,y)$ $\displaystyle f_{xx}(x,y)$.

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=(12x^2)(12y^2)-(3)^2=144x^2y^2-9$

At (0,0) $\displaystyle (0,0)$,

D(x,y)=144(0)^2(0)^2-9=-9<0 $\displaystyle D(x,y)=144(0)^2(0)^2-9=-9< 0$

Since D(x,y)<0 $\displaystyle D(x,y)< 0$, (0,0) $\displaystyle (0,0)$ is a saddle point.

At $(-\sqrt3/2,\sqrt3/2)$ $\displaystyle (-\sqrt3/2,\sqrt3/2)$,

$D(x,y)=144\left ( \frac{-\sqrt3}{2} \right )^2\left ( \frac{\sqrt3}{2} \right )^2-9$ $\displaystyle D(x,y)=144\left ( \frac{-\sqrt3}{2} \right )^2\left ( \frac{\sqrt3}{2} \right )^2-9$

$D(x,y)=144\left ( \frac{3}{4} \right )\left ( \frac{3}{4} \right )-9=72>0$ $\displaystyle D(x,y)=144\left ( \frac{3}{4} \right )\left ( \frac{3}{4} \right )-9=72>0$

$f_{xx}(x,y)=12x^2=12\left ( \frac{-\sqrt3}{2} \right )^2=12\left ( \frac{3}{4} \right )=9>0$ $\displaystyle f_{xx}(x,y)=12x^2=12\left ( \frac{-\sqrt3}{2} \right )^2=12\left ( \frac{3}{4} \right )=9>0$

Since $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, $(-\sqrt3/2,\sqrt3/2)$ $\displaystyle (-\sqrt3/2,\sqrt3/2)$ is a minimum.

At $(\sqrt3/2,-\sqrt3/2)$ $\displaystyle (\sqrt3/2,-\sqrt3/2)$,

$D(x,y)=144\left ( \frac{\sqrt3}{2} \right )^2\left ( \frac{-\sqrt3}{2} \right )^2-9$ $\displaystyle D(x,y)=144\left ( \frac{\sqrt3}{2} \right )^2\left ( \frac{-\sqrt3}{2} \right )^2-9$

$D(x,y)=144\left ( \frac{3}{4} \right )\left ( \frac{3}{4} \right )-9=72>0$ $\displaystyle D(x,y)=144\left ( \frac{3}{4} \right )\left ( \frac{3}{4} \right )-9=72>0$

$f_{xx}(x,y)=12x^2=12\left ( \frac{\sqrt3}{2} \right )^2=12\left ( \frac{3}{4} \right )=9>0$ $\displaystyle f_{xx}(x,y)=12x^2=12\left ( \frac{\sqrt3}{2} \right )^2=12\left ( \frac{3}{4} \right )=9>0$

Since $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, $(-\sqrt3/2,\sqrt3/2)$ $\displaystyle (-\sqrt3/2,\sqrt3/2)$ is a minimum.

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Example Question #8 : Relative Minimums And Maximums

Find the relative maxima and minima of $\displaystyle f(x,y)=sin(x)-cos(y)+4xy$.

Possible Answers:

$\displaystyle (0.0617,-0.2495)$ is a relative maxima

(0,0) $\displaystyle (0,0)$ and $\displaystyle (0.0617,-0.2495)$ is a relative minima

(0,0) $\displaystyle (0,0)$ and $\displaystyle (0.0617,-0.2495)$ is a relative maxima

$\displaystyle (0.0617,-0.2495)$ is a saddle point

Correct answer:

$\displaystyle (0.0617,-0.2495)$ is a saddle point

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)<0$ $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If D(x,y)<0 $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

$\displaystyle f(x,y)=sin(x)-cos(y)+4xy$

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=cos(x)+4y$ $\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=cos(x)+4y$

$\frac{\partial f }{\partial y}=f_y(x,y)=sin(y)+4x$ $\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=sin(y)+4x$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=-sin(x)$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=-sin(x)$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=cos(y)$ $\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=cos(y)$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$f_x(x,y)=cos(x)+4y=0\Rightarrow cos(x)=-4y$ $\displaystyle f_x(x,y)=cos(x)+4y=0\Rightarrow cos(x)=-4y$

$\displaystyle y=-cos(x)/4$

$f_y(x,y)=sin(y)+4x=0\Rightarrow sin\left (\frac{-cos(x)}{4} \right )+4x=0$ $\displaystyle f_y(x,y)=sin(y)+4x=0\Rightarrow sin\left (\frac{-cos(x)}{4} \right )+4x=0$

Using a TI-83 or other software to find the root, we find that $x\approx 0.0617$ $\displaystyle x\approx 0.0617$,

We find the corresponding value of $\displaystyle y$ using $\displaystyle y=-cos(x)/4$ (found by rearranging the first derivative)

$y(0.0617)=-cos(0.0617)/4 \approx-0.2495$ $\displaystyle y(0.0617)=-cos(0.0617)/4 \approx-0.2495$

There is a critical points at $\displaystyle (0.0617,-0.2495)$. We need to determine if the critical point is a maximum or minimum using D(x,y) $\displaystyle D(x,y)$ and $f_{xx}(x,y)$ $\displaystyle f_{xx}(x,y)$.

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=-sin(x)cos(y)-4^2$

At $\displaystyle (0.0617,-0.2495)$,

$D(x,y)=-sin(0.0617)cos(-0.2495)-4^2\approx -16.06<0$ $\displaystyle D(x,y)=-sin(0.0617)cos(-0.2495)-4^2\approx -16.06< 0$

Since D(x,y)<0 $\displaystyle D(x,y)< 0$, $\displaystyle (0.0617,-0.2495)$ is a saddle point.

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Example Question #9 : Relative Minimums And Maximums

Find the relative maxima and minima of $\displaystyle f(x,y)=sin^2x+cos^2y-16$.

Possible Answers:

(0,0) $\displaystyle (0,0)$, $(0,3\pi/2)$ $\displaystyle (0,3\pi/2)$, $(\pi/2,\pi/2)$ $\displaystyle (\pi/2,\pi/2)$, $(\pi/2,\pi)$ $\displaystyle (\pi/2,\pi)$, $(\pi,\pi/2)$ $\displaystyle (\pi,\pi/2)$, $(\pi,\pi)$ $\displaystyle (\pi,\pi)$, $(3\pi/2,0)$ $\displaystyle (3\pi/2,0)$and $(3\pi/2,3\pi/2)$ $\displaystyle (3\pi/2,3\pi/2)$ are relative maxima

$(0,\pi/2)$ $\displaystyle (0,\pi/2)$, $(0,\pi)$ $\displaystyle (0,\pi)$, $(3\pi/2,\pi/2)$ $\displaystyle (3\pi/2,\pi/2)$ and $(3\pi/2,\pi)$ $\displaystyle (3\pi/2,\pi)$, $(\pi/2,0)$ $\displaystyle (\pi/2,0)$, $(\pi/2,3\pi/2)$ $\displaystyle (\pi/2,3\pi/2)$, $(\pi,0)$ $\displaystyle (\pi,0)$ and $(\pi,3\pi/2)$ $\displaystyle (\pi,3\pi/2)$ are saddle points

$(0,\pi/2)$ $\displaystyle (0,\pi/2)$, $(0,\pi)$ $\displaystyle (0,\pi)$, $(3\pi/2,\pi/2)$ $\displaystyle (3\pi/2,\pi/2)$ and $(3\pi/2,\pi)$ $\displaystyle (3\pi/2,\pi)$are saddle points

$(\pi/2,0)$ $\displaystyle (\pi/2,0)$, $(\pi/2,3\pi/2)$ $\displaystyle (\pi/2,3\pi/2)$, $(\pi,0)$ $\displaystyle (\pi,0)$ and $(\pi,3\pi/2)$ $\displaystyle (\pi,3\pi/2)$ are relative minima

$(0,\pi/2)$ $\displaystyle (0,\pi/2)$, $(0,\pi)$ $\displaystyle (0,\pi)$, $(3\pi/2,\pi/2)$ $\displaystyle (3\pi/2,\pi/2)$ and $(3\pi/2,\pi)$ $\displaystyle (3\pi/2,\pi)$are relative minima

$(\pi/2,0)$ $\displaystyle (\pi/2,0)$, $(\pi/2,3\pi/2)$ $\displaystyle (\pi/2,3\pi/2)$, $(\pi,0)$ $\displaystyle (\pi,0)$ and $(\pi,3\pi/2)$ $\displaystyle (\pi,3\pi/2)$ are relative maxima

$(0,\pi/2)$ $\displaystyle (0,\pi/2)$, $(0,\pi)$ $\displaystyle (0,\pi)$, $(3\pi/2,\pi/2)$ $\displaystyle (3\pi/2,\pi/2)$ and $(3\pi/2,\pi)$ $\displaystyle (3\pi/2,\pi)$are relative maxima

$(\pi/2,0)$ $\displaystyle (\pi/2,0)$, $(\pi/2,3\pi/2)$ $\displaystyle (\pi/2,3\pi/2)$, $(\pi,0)$ $\displaystyle (\pi,0)$ and $(\pi,3\pi/2)$ $\displaystyle (\pi,3\pi/2)$ are saddle points

Correct answer:

$(0,\pi/2)$ $\displaystyle (0,\pi/2)$, $(0,\pi)$ $\displaystyle (0,\pi)$, $(3\pi/2,\pi/2)$ $\displaystyle (3\pi/2,\pi/2)$ and $(3\pi/2,\pi)$ $\displaystyle (3\pi/2,\pi)$are relative minima

$(\pi/2,0)$ $\displaystyle (\pi/2,0)$, $(\pi/2,3\pi/2)$ $\displaystyle (\pi/2,3\pi/2)$, $(\pi,0)$ $\displaystyle (\pi,0)$ and $(\pi,3\pi/2)$ $\displaystyle (\pi,3\pi/2)$ are relative maxima

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)<0$ $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If D(x,y)<0 $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=2sin(x)cos(x)$ $\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=2sin(x)cos(x)$

$\frac{\partial f }{\partial y}=f_y(x,y)=2cos(y)(-sin(y))=-2sin(y)cos(y)$ $\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=2cos(y)(-sin(y))=-2sin(y)cos(y)$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2\left [ sin(x)(-sin(x))+cos(x)cos(x)\right ]$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2\left [ sin(x)(-sin(x))+cos(x)cos(x)\right ]$

$f_{xx}(x,y)=2\left [ -sin^2(x)+cos^2(x)\right ]=2cos^2(x)-2sin^2(x)$ $\displaystyle f_{xx}(x,y)=2\left [ -sin^2(x)+cos^2(x)\right ]=2cos^2(x)-2sin^2(x)$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=-2\left [ sin(y)(-sin(y))+cos(y)cos(y)\right ]$ $\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=-2\left [ sin(y)(-sin(y))+cos(y)cos(y)\right ]$

$f_{yy}(x,y)=-2\left [ -sin^2(y)+cos^2(y)\right ]=2sin^2(y)-2cos^2(y)$ $\displaystyle f_{yy}(x,y)=-2\left [ -sin^2(y)+cos^2(y)\right ]=2sin^2(y)-2cos^2(y)$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=0$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=0$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$\displaystyle f_x(x,y)=2sin(x)cos(x)=0$

$sin(x)=0\Rightarrow x=0, \pi$ $\displaystyle sin(x)=0\Rightarrow x=0, \pi$

$cos(x)=0\Rightarrow x= \pi/2, 3\pi/2$ $\displaystyle cos(x)=0\Rightarrow x= \pi/2, 3\pi/2$

$\displaystyle f_y(x,y)=-2sin(y)cos(y)=0$

$sin(y)=0\Rightarrow y=0, \pi$ $\displaystyle sin(y)=0\Rightarrow y=0, \pi$

$cos(y)=0\Rightarrow y= \pi/2, 3\pi/2$ $\displaystyle cos(y)=0\Rightarrow y= \pi/2, 3\pi/2$

Our derivatives equal $\displaystyle 0$ when $x=0, \pi/2, \pi, 3\pi/2$ $\displaystyle x=0, \pi/2, \pi, 3\pi/2$ and $y=0, \pi/2, \pi, 3\pi/2$ $\displaystyle y=0, \pi/2, \pi, 3\pi/2$. Every linear combination of these points is a critical point. The critical points are

(0,0) $\displaystyle (0,0)$, $(0,\pi/2)$ $\displaystyle (0,\pi/2)$, $(0,\pi)$ $\displaystyle (0,\pi)$, $(0,3\pi/2)$ $\displaystyle (0,3\pi/2)$

$(\pi/2,0)$ $\displaystyle (\pi/2,0)$, $(\pi/2,\pi/2)$ $\displaystyle (\pi/2,\pi/2)$, $(\pi/2,\pi)$ $\displaystyle (\pi/2,\pi)$, $(\pi/2,3\pi/2)$ $\displaystyle (\pi/2,3\pi/2)$

$(\pi,0)$ $\displaystyle (\pi,0)$, $(\pi,\pi/2)$ $\displaystyle (\pi,\pi/2)$, $(\pi,\pi)$ $\displaystyle (\pi,\pi)$, $(\pi,3\pi/2)$ $\displaystyle (\pi,3\pi/2)$

$(3\pi/2,0)$ $\displaystyle (3\pi/2,0)$, $(3\pi/2,\pi/2)$ $\displaystyle (3\pi/2,\pi/2)$, $(3\pi/2,\pi)$ $\displaystyle (3\pi/2,\pi)$, $(3\pi/2,3\pi/2)$ $\displaystyle (3\pi/2,3\pi/2)$

We need to determine if the critical point is a maximum or minimum using D(x,y) $\displaystyle D(x,y)$ and $f_{xx}(x,y)$ $\displaystyle f_{xx}(x,y)$.

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=(2cos^2x-2sin^2x)(2sin^2y-2cos^2y)-0^2$

$f_{xx}(x,y)=2cos^2(x)-2sin^2(x)$ $\displaystyle f_{xx}(x,y)=2cos^2(x)-2sin^2(x)$

(0,0) $\displaystyle (0,0)$, $(0,\pi/2)$ $\displaystyle (0,\pi/2)$, $(0,\pi)$ $\displaystyle (0,\pi)$, $(0,3\pi/2)$ $\displaystyle (0,3\pi/2)$

$\displaystyle D(0,0)=(2cos^2(0)-2sin^2(0))(2sin^2(0)-2cos^2(0))-0^2$

$\displaystyle =(2(1)-0)(2(0)-2(1))=(2)(-2)=-4< 0$

Saddle point

$D(0,\pi/2)=(2cos^2(0)-2sin^2(0))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$ $\displaystyle D(0,\pi/2)=(2cos^2(0)-2sin^2(0))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$

$\displaystyle =(2(1)-2(0))(2(1)-2(0))=(2)(2)=4>0$

$f_{xx}(0,\pi/2)=2cos^2(0)-2sin^2(0)=2(1)-2(0)=2>0$ $\displaystyle f_{xx}(0,\pi/2)=2cos^2(0)-2sin^2(0)=2(1)-2(0)=2>0$

minimum

$D(0,\pi)=(2cos^2(0)-2sin^2(0))(2sin^2(\pi)-2cos^2(\pi))-0^2$ $\displaystyle D(0,\pi)=(2cos^2(0)-2sin^2(0))(2sin^2(\pi)-2cos^2(\pi))-0^2$

$\displaystyle =(2(1)-0)(2(0)-2(-1))=(2)(2)=4>0$

$f_{xx}(0,\pi)=2cos^2(0)-2sin^2(0)=2(1)-2(0)=2>0$ $\displaystyle f_{xx}(0,\pi)=2cos^2(0)-2sin^2(0)=2(1)-2(0)=2>0$

minimum

$D(0,3\pi/2)=(2cos^2(0)-2sin^2(0))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$ $\displaystyle D(0,3\pi/2)=(2cos^2(0)-2sin^2(0))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$

$\displaystyle =(2(1)-2(0))(2(-1)-2(0))=(2)(-2)=-4< 0$

Saddle point

$(\pi/2,0)$ $\displaystyle (\pi/2,0)$, $(\pi/2,\pi/2)$ $\displaystyle (\pi/2,\pi/2)$, $(\pi/2,\pi)$ $\displaystyle (\pi/2,\pi)$, $(\pi/2,3\pi/2)$ $\displaystyle (\pi/2,3\pi/2)$

$D(\pi/2,0)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(0)-2cos^2(0))-0^2$ $\displaystyle D(\pi/2,0)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(0)-2cos^2(0))-0^2$

$\displaystyle =(2(0)-2(1))(2(0)-2(1))=(-2)(-2)=4>0$

$f_{xx}(\pi/2,0)=2cos^2(\pi/2)-2sin^2(\pi/2)=2(0)-2(1)=-2<0$ $\displaystyle f_{xx}(\pi/2,0)=2cos^2(\pi/2)-2sin^2(\pi/2)=2(0)-2(1)=-2< 0$

maximum

$D(\pi/2,\pi/2)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$ $\displaystyle D(\pi/2,\pi/2)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$

$\displaystyle =(2(0)-2(1))(2(1)-2(0))=(-2)(2)=-4< 0$

saddle point

$D(\pi/2,\pi)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(\pi)-2cos^2(\pi))-0^2$ $\displaystyle D(\pi/2,\pi)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(\pi)-2cos^2(\pi))-0^2$

$\displaystyle =(2(0)-2(1))(2(0)-2(-1))=(-2)(2)=-4< 0$

saddle point

$D(\pi/2,3\pi/2)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$ $\displaystyle D(\pi/2,3\pi/2)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$

$\displaystyle =(2(0)-2(1))(2(-1)-2(0))=(-2)(-2)=4>0$

$f_{xx}(\pi/2,2\pi/2)=2cos^2(\pi/2)-2sin^2(\pi/2)=2(0)-2(1)=-2<0$ $\displaystyle f_{xx}(\pi/2,2\pi/2)=2cos^2(\pi/2)-2sin^2(\pi/2)=2(0)-2(1)=-2< 0$

maximum

$(\pi,0)$ $\displaystyle (\pi,0)$, $(\pi,\pi/2)$ $\displaystyle (\pi,\pi/2)$, $(\pi,\pi)$ $\displaystyle (\pi,\pi)$, $(\pi,3\pi/2)$ $\displaystyle (\pi,3\pi/2)$

$D(\pi,0)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(0)-2cos^2(0))-0^2$ $\displaystyle D(\pi,0)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(0)-2cos^2(0))-0^2$

$\displaystyle =(2(-1)-0)(2(0)-2(1))=(-2)(-2)=4>0$

$f_{xx}(\pi,\pi/2)=2cos^2(\pi)-2sin^2(\pi)=2(-1)-2(0)=-2<0$ $\displaystyle f_{xx}(\pi,\pi/2)=2cos^2(\pi)-2sin^2(\pi)=2(-1)-2(0)=-2< 0$

maximum

$D(\pi,\pi/2)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$ $\displaystyle D(\pi,\pi/2)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$

$\displaystyle =(2(-1)-2(0))(2(1)-2(0))=(-2)(2)=-4< 0$

saddle point

$D(\pi,\pi)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(\pi)-2cos^2(\pi))-0^2$ $\displaystyle D(\pi,\pi)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(\pi)-2cos^2(\pi))-0^2$

$\displaystyle =(2(-1)-0)(2(0)-2(-1))=(-2)(2)=-4< 0$

saddle point

$D(\pi,3\pi/2)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$ $\displaystyle D(\pi,3\pi/2)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$

$\displaystyle =(2(-1)-2(0))(2(-1)-2(0))=(-2)(-2)=4>0$

$f_{xx}(\pi,3\pi/2)=2cos^2(\pi)-2sin^2(\pi)=2(-1)-2(0)=-2<0$ $\displaystyle f_{xx}(\pi,3\pi/2)=2cos^2(\pi)-2sin^2(\pi)=2(-1)-2(0)=-2< 0$

maximum

$(3\pi/2,0)$ $\displaystyle (3\pi/2,0)$, $(3\pi/2,\pi/2)$ $\displaystyle (3\pi/2,\pi/2)$, $(3\pi/2,\pi)$ $\displaystyle (3\pi/2,\pi)$, $(3\pi/2,3\pi/2)$ $\displaystyle (3\pi/2,3\pi/2)$

$D(3\pi/2,0)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(0)-2cos^2(0))-0^2$ $\displaystyle D(3\pi/2,0)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(0)-2cos^2(0))-0^2$

$\displaystyle =(2(0)-2(-1))(2(0)-2(1))=(2)(-2)=-4< 0$

saddle point

$D(3\pi/2,\pi/2)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$ $\displaystyle D(3\pi/2,\pi/2)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$

$\displaystyle =(2(0)-2(-1))(2(1)-2(0))=(2)(2)=4>0$

$f_{xx}(3\pi/2,\pi/2)=2cos^2(3\pi/2)-2sin^2(3\pi/2)=2(0)-2(-1)=2>0$ $\displaystyle f_{xx}(3\pi/2,\pi/2)=2cos^2(3\pi/2)-2sin^2(3\pi/2)=2(0)-2(-1)=2>0$

minimum

$D(3\pi/2,\pi)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(\pi)-2cos^2(\pi))-0^2$ $\displaystyle D(3\pi/2,\pi)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(\pi)-2cos^2(\pi))-0^2$

$\displaystyle =(2(0)-2(-1))(2(0)-2(-1))=(2)(2)=4>0$

$f_{xx}(3\pi/2,\pi)=2cos^2(3\pi/2)-2sin^2(3\pi/2)=2(0)-2(-1)=2>0$ $\displaystyle f_{xx}(3\pi/2,\pi)=2cos^2(3\pi/2)-2sin^2(3\pi/2)=2(0)-2(-1)=2>0$

minimum

$D(3\pi/2,3\pi/2)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$ $\displaystyle D(3\pi/2,3\pi/2)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$

$\displaystyle =(2(0)-2(-1))(2(-1)-2(0))=(2)(-2)=-4< 0$

saddle point

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Example Question #10 : Relative Minimums And Maximums

Find the relative maxima and minima of $\displaystyle f(x,y)=x^2+y^2$.

Possible Answers:

(0,0) $\displaystyle (0,0)$ is a relative minimum

(1,1) $\displaystyle (1,1)$ is a relative minimum

(0,0) $\displaystyle (0,0)$ is a relative maximum

(1,1) $\displaystyle (1,1)$ is a relative maximum

Correct answer:

(0,0) $\displaystyle (0,0)$ is a relative minimum

Explanation:

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)<0$ $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If D(x,y)<0 $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=2x$ $\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=2x$

$\frac{\partial f }{\partial y}=f_y(x,y)=2y$ $\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=2y$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2$ $\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=2$ $\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=2$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$f_x(x,y)=2x=0\Rightarrow x=0$ $\displaystyle f_x(x,y)=2x=0\Rightarrow x=0$

$f_y(x,y)=2y=0\Rightarrow y=0$ $\displaystyle f_y(x,y)=2y=0\Rightarrow y=0$

There is a critical point at (0,0) $\displaystyle (0,0)$. We need to determine if the critical point is a maximum or minimum using D(x,y) $\displaystyle D(x,y)$ and $f_{xx}(x,y)$ $\displaystyle f_{xx}(x,y)$.

$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$ $\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=2(2)-0^2=4$

At (0,0) $\displaystyle (0,0)$,

$\displaystyle D(x,y)=4>0$

$f_{xx}(x,y)=2>0$ $\displaystyle f_{xx}(x,y)=2>0$

Since $\displaystyle D(x,y)>0$ and $f_{xx}(x,y)>0$ $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at (0,0) $\displaystyle (0,0)$.

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Calculus 3 : Relative Minimums and Maximums

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #121 : Applications Of Partial Derivatives

Example Question #2 : Relative Minimums And Maximums

Example Question #3 : Relative Minimums And Maximums

Example Question #2 : Relative Minimums And Maximums

Example Question #5 : Relative Minimums And Maximums

Example Question #6 : Relative Minimums And Maximums

Example Question #3 : Relative Minimums And Maximums

Example Question #8 : Relative Minimums And Maximums

Example Question #9 : Relative Minimums And Maximums

Example Question #10 : Relative Minimums And Maximums

All Calculus 3 Resources

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