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Calculus 3 : Relative Minimums and Maximums

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Example Questions

Example Question #11 : Relative Minimums And Maximums

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

Possible Answers:

(-1,0) $\displaystyle (-1,0)$ and (1,0) $\displaystyle (1,0)$ are saddle points

(-1,0) $\displaystyle (-1,0)$ is a relative minima and (1,0) $\displaystyle (1,0)$ is a relative maxima

(-1,0) $\displaystyle (-1,0)$ and (1,0) $\displaystyle (1,0)$ are relative minima

(-1,0) $\displaystyle (-1,0)$ is a relative maxima and (1,0) $\displaystyle (1,0)$ is a relative minima

Correct answer:

(-1,0) $\displaystyle (-1,0)$ and (1,0) $\displaystyle (1,0)$ are saddle points

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.

$\displaystyle f(x,y)=3x^2y^2-3x^2$

The first order partial derivatives are

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

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

The second order partial derivatives are

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

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

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

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

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

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

The critical points are (-1,0) $\displaystyle (-1,0)$ and (1,0) $\displaystyle (1,0)$. We need to determine if the critical point is a maximum or minimum using D(x,y) $\displaystyle D(x,y)$ and $f_{xx}(x,y)$ $\displaystyle f_{xx}(x,y)$.

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

$\displaystyle D(x,y)=(6y^2-6)(6x^2)-(12xy)^2=36x^2y^2-36x^2-144x^2y^2$

$\displaystyle D(x,y)=-108x^2y^2-36x^2$

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

$\displaystyle D(x,y)=-108(-1)^2(0)^2-36(-1)^2=-36< 0$

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

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

$\displaystyle D(x,y)=-108(-1)^2(0)^2-36(1)^2=-36< 0$

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

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Example Question #132 : Applications Of Partial Derivatives

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

Possible Answers:

(0,0) $\displaystyle (0,0)$ is a saddle point

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

(5/4,0) $\displaystyle (5/4,0)$ is a relative minima

(5/4,0) $\displaystyle (5/4,0)$ is a saddle point

Correct answer:

(5/4,0) $\displaystyle (5/4,0)$ 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

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

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

The second order partial derivatives are

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

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

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

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

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

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

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

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

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

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

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

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

The exponential part of each expression cannot equal $\displaystyle 0$, so each derivative is $\displaystyle 0$ only when $\displaystyle (15x^2-2)=0$ and 10y=0 $\displaystyle 10y=0$. That is $x=\pm \sqrt{2/15}$ $\displaystyle x=\pm \sqrt{2/15}$ and y=0 $\displaystyle y=0$.

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

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

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

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

At (5/4,0) $\displaystyle (5/4,0)$,

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

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

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

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

Since D(x,y)<0 $\displaystyle D(x,y)< 0$, (5/4,0) $\displaystyle (5/4,0)$ is a saddle point.

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Example Question #13 : Relative Minimums And Maximums

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

Possible Answers:

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

(0,0) $\displaystyle (0,0)$ is a saddle point

(5,-7) $\displaystyle (5,-7)$ is a relative maximum

(5,-7) $\displaystyle (5,-7)$ is a saddle point

Correct answer:

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

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

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

The second order partial derivatives are

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

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

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

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

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

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

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

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

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

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

The exponential part of each expression cannot equal $\displaystyle 0$, so each derivative is $\displaystyle 0$ only when $\displaystyle (10x-7y)=0$ and -7x=0 $\displaystyle -7x=0$. That is x=0 $\displaystyle x=0$ and y=0 $\displaystyle y=0$.

The critical point is (0,0) $\displaystyle (0,0)$. We need to determine if the critical points are maxima or minima using D(x,y) $\displaystyle D(x,y)$ and $f_{xx}(x,y)$ $\displaystyle f_{xx}(x,y)$.

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

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

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

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

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

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

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

D(x,y)=(-49)(1)^2<0 $\displaystyle D(x,y)=(-49)(1)^2< 0$

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

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Calculus 3 : Relative Minimums and Maximums

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Example Question #11 : Relative Minimums And Maximums

Example Question #132 : Applications Of Partial Derivatives

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