Relative Minimums and Maximums - AP Calculus BC

Card 0 of 65

Question

Find and classify all the critical points for .

Answer

First thing we need to do is take partial derivatives.

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

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

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

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

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

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

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

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

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

Lets summarize the critical points:

If

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

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

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

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

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

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

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

If , there is a saddle point at

If then the point may be a relative minimum, relative maximum or a saddle point.

Now we plug in the critical values into .

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

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

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

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

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

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

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

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

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

We find the corresponding values of using (found by rearranging the first derivative)

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

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

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

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

At ,

Since , is a saddle point.

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

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

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

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

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

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

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

Squaring both sides of the equation gives us

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

Multiplying both sides of the equation by gives us

There are three possible values of ; , and .

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

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

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

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

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

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

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

At ,

Since , is a saddle point.

At ,

Since , is a saddle point.

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

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

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

Since , $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ is a saddle point.

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

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

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

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

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

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

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

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

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

At ,

Since , is a saddle point.

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

The real values of are and

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

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

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

There are critical points at and . We need to determine if the critical points are maxima or minima using and $f_{xx}(x,y)$ .

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

At ,

Since , is a saddle point.

At ,

Since , is a saddle point.

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

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

Setting each factor in the expression equal to gives us

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

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

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

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

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

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

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

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

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

At ,

Since , is a saddle point.

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

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

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

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

We find the corresponding value of using (found by rearranging the first derivative)

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

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

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

At ,

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

Since , is a saddle point.

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

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

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

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

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

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

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

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

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

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

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

Saddle point

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

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

minimum

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

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

minimum

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

Saddle point

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

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

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

maximum

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

saddle point

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

saddle point

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

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

maximum

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

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

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

maximum

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

saddle point

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

saddle point

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

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

maximum

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

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

saddle point

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

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

minimum

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

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

minimum

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

saddle point

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

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

At ,

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

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

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

$\frac{\partial f }{\partial y}=f_y(x,y)=6x^2y$

The second order partial derivatives are

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

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=6x^2$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=12xy$

To find the critical points, we will set the first derivatives equal to

$f_{xx}(x,y)=6y^2-6=0\Rightarrow 6y^2=6\Rightarrow y=\pm1$

$f_{yy}(x,y)=6x^2=0 \Rightarrow x=0$

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

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

At ,

Since , is a saddle point.

At ,

Since , is a saddle point.

Compare your answer with the correct one above

Question

Find the relative maxima and minima of $f(x,y)=e^{5x^3+5y^2-2x}$ .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=e^{5x^3+5y^2-2x}(15x^2-2)=(15x^2-2)e^{5x^3+5y^2-2x}$

$\frac{\partial f }{\partial y}=f_y(x,y)=e^{5x^3+5y^2-2x}(10y)=(10y)e^{5x^3+5y^2-2x}$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=(15x^2-2)e^{5x^3+5y^2-2x}(15x^2-2)+e^{5x^3+5y^2-2x}(30x)$

$f_{xx}(x,y)=((15x^2-2)^2+30x)e^{5x^3+5y^2-2x}$

$f_{xx}(x,y)=(225x^4-60x^2+30x+4)e^{5x^3+5y^2-2x}$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=(10y)e^{5x^3+5y^2-2x}(10y)+e^{5x^3+5y^2-2x}(10)$

$f_{yy}(x,y)=(100y^2)e^{5x^3+5y^2-2x}+(10)e^{5x^3+5y^2-2x}$

$f_{yy}(x,y)=(100y^2+10)e^{5x^3+5y^2-2x}$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=(100y^2+10)e^{5x^3+5y^2-2x}(15x^2-2)$

$f_{xy}(x,y)=(100y^2+10)(15x^2-2)e^{5x^3+5y^2-2x}$

$f_{xy}(x,y)=(1500x^2y^2+150x^2-200y^2-20)e^{5x^3+5y^2-2x}$

To find the critical points, we will set the first derivatives equal to

$f_x(x,y)=(15x^2-2)e^{5x^3+5y^2-2x}$

$f_y(x,y)=(10y)e^{5x^3+5y^2-2x}$

The exponential part of each expression cannot equal , so each derivative is only when and . That is $x=\pm \sqrt{2/15}$ and .

The critical points are $(-\sqrt{2/15},0)$ and $(\sqrt{2/15},0)$ . We need to determine if the critical points are maxima or minima using and $f_{xx}(x,y)$ .

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

$D(x,y)=(225x^4-60x^2+30x+4)e^{5x^3+5y^2-2x}(100y^2+10)e^{5x^3+5y^2-2x} -((1500x^2y^2+150x^2-200y^2-20)e^{5x^3+5y^2-2x})^2$

$D(x,y)=(225x^4-60x^2+30x+4)(100y^2+10)(e^{5x^3+5y^2-2x})^2 -(1500x^2y^2+150x^2-200y^2-20)^2(e^{5x^3+5y^2-2x})^2$

At ,

$D(x,y)=(225(5/4)^4-60(5/4)^2+30(5/4)+4)(0+10)(e^{5(5/4)^3+5(0)^2-2(5/4)})^2 -(1500(5/4)^2(0)^2+150(5/4)^2-200(0)^2-20)^2(e^{5(5/4)^3+5(0)^2-2(5/4)})^2$

$D(x,y)=(225(5/4)^4-60(5/4)^2+30(5/4)+4)(10)(e^{5(5/4)^3-2(5/4)})^2 -(150(5/4)^2-20)^2(e^{5(5/4)^3-2(5/4)})^2$

$D(x,y)\approx(497)(10)(2.0457) -(45956.6)(2.0457)$

$D(x,y)\approx -83846.3<0$

Since , is a saddle point.

Compare your answer with the correct one above

Question

Find the relative maxima and minima of $f(x,y)=e^{5x^2-7xy}$ .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

$\frac{\partial f }{\partial x}=f_x(x,y)=e^{5x^2-7xy}(10x-7y)$

$\frac{\partial f }{\partial y}=f_y(x,y)=e^{5x^2-7xy}(-7x)$

The second order partial derivatives are

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=e^{5x^2-7xy}(10)+(10x-7y)e^{5x^2-7xy}(10x-7y)$

$f_{xx}(x,y)=(10+(10x-7y)^2)e^{5x^2-7xy}$

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=e^{5x^2-7xy}(-7x)(-7x)$

$f_{yy}(x,y)=90x^2e^{5x^2-7xy}$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=e^{5x^2-7xy}(-7)+(-7x)e^{5x^2-7xy}(10x-7y)$

$f_{xy}(x,y)=(-7+(-7x)(10x-7y))e^{5x^2-7xy}$

$f_{xy}(x,y)=(-7-70x^2+49xy)e^{5x^2-7xy}$

To find the critical points, we will set the first derivatives equal to

$\frac{\partial f }{\partial x}=f_x(x,y)=e^{5x^2-7xy}(10x-7y)$

$\frac{\partial f }{\partial y}=f_y(x,y)=e^{5x^2-7xy}(-7x)$

The exponential part of each expression cannot equal , so each derivative is only when and . That is and .

The critical point is . We need to determine if the critical points are maxima or minima using and $f_{xx}(x,y)$ .

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

$D(x,y)=(10+(10x-7y)^2)e^{5x^2-7xy}90x^2*e^{5x^2-7xy}-((-7-70x^2+49xy)e^{5x^2-7xy})^2$

$D(x,y)=(90x^2)(10+(10x-7y)^2)(e^{5x^2-7xy})^2 -(-7-70x^2+49xy)^2(e^{5x^2-7xy})^2$

$D(x,y)=\left [(90x^2)(10+(10x-7y)^2)-(-7-70x^2+49xy)^2 \right ](e^{5x^2-7xy})^2$

$D(x,y)=(4100x^4-5740x^3y+2009x^2y^2-80x^2+686xy-49)(e^{5x^2-7xy})^2$

At ,

$D(x,y)=(4100(0)-5740(0)+2009(0)-80(0)+686(0)-49)(e^{5(0)-7(0)})^2$

Since , is a saddle point.

Compare your answer with the correct one above

Question

Find and classify all the critical points for .

Answer

First thing we need to do is take partial derivatives.

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

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

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

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

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

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

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

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

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

Lets summarize the critical points:

If

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

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

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

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

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

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

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

If , there is a saddle point at

If then the point may be a relative minimum, relative maximum or a saddle point.

Now we plug in the critical values into .

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

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

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

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

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

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

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

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

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

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

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

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

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

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

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

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

We find the corresponding values of using (found by rearranging the first derivative)

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

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

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

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

At ,

Since , is a saddle point.

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

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

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

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

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

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

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

Squaring both sides of the equation gives us

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

Multiplying both sides of the equation by gives us

There are three possible values of ; , and .

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

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

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

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

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

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

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

At ,

Since , is a saddle point.

At ,

Since , is a saddle point.

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

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

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

Since , $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ is a saddle point.

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

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

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

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

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

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

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

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

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

At ,

Since , is a saddle point.

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

$\frac{\partial f }{\partial y}=f_y(x,y)=6x^2y$

The second order partial derivatives are

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

$\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=6x^2$

$\frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=12xy$

To find the critical points, we will set the first derivatives equal to

$f_{xx}(x,y)=6y^2-6=0\Rightarrow 6y^2=6\Rightarrow y=\pm1$

$f_{yy}(x,y)=6x^2=0 \Rightarrow x=0$

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

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

At ,

Since , is a saddle point.

At ,

Since , is a saddle point.

Compare your answer with the correct one above

Question

Find the relative maxima and minima of .

Answer

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

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

to classify the critical points.

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

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

If , then this point is a saddle point.

If , then this point cannot be classified.

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

To find the critical points, we will set the first derivatives equal to

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

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

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

The real values of are and

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

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

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

There are critical points at and . We need to determine if the critical points are maxima or minima using and $f_{xx}(x,y)$ .

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

At ,

Since , is a saddle point.

At ,

Since , is a saddle point.

Compare your answer with the correct one above

Tap the card to reveal the answer