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Calculus 3 : Relative Minimums and Maximums

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Example Question #1 : Relative Minimums And Maximums

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

Possible Answers:

(0,0): Relative Minimum

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

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

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

(0,0): Saddle Point

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

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

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

(0,0): Relative Maximum

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

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

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

(0,0): Relative Minimum

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

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

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

(0,0): Relative Minimum

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

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

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

Correct answer:

(0,0): Relative Minimum

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

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

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

Explanation:

First thing we need to do is take partial derivatives.

f_x(x,y)=10xy+4x

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

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

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

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

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.

x=0:

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

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

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

Lets summarize the critical points:

If x=0:

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

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 a,b , is a critical point.

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

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

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

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

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

Now we plug in the critical values into .

(0,0):

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

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

Since D>0 and $f_{xx}(a,b)>0$ , (0,0) 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 D<0 , $(0,\frac{1}{2})$ 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 D<0 , $\Big(\sqrt{\frac{108}{125}},-\frac{2}{5}\Big)$ 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 D<0 , $\Big(-\sqrt{\frac{108}{125}},-\frac{2}{5}\Big)$ is a saddle point

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Example Question #2 : Relative Minimums And Maximums

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

Possible Answers:

(0,0) is a relative maximum.

(5,5) is a relative maximum.

(5,5) is a relative minimum.

(0,0) is a relative minimum.

Correct answer:

(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

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

to classify the critical points.

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

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

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

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

The first order partial derivatives are

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

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

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

f_x(x,y)=10x+4y=0

f_y(x,y)=10y+4x=0

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

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

84y/10=0

y=0

x=-4y/10=-4(0)/10=0

There is only one critical point and it is at (0,0) . We need to determine if this critical point is a maximum or minimum using D(x,y) 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(10)-(4)^2=100-16=84

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

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

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Example Question #3 : Relative Minimums And Maximums

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

to classify the critical points.

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

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

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

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

The first order partial derivatives are

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

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

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)=12y^2-2x=0

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

x(27x^3-2)=0

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

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

y(0)=3(0)^2/2=0

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

There are critical points at (0,0) 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 D(x,y) 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 (0,0) ,

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

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

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Example Question #4 : Relative Minimums And Maximums

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

Possible Answers:

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

(0,0) , (-1,0) and $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ are saddle points.

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

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

Correct answer:

(0,0) , (-1,0) and $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ are saddle points.

Explanation:

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

to classify the critical points.

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

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

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

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

The first order partial derivatives are

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

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

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

-x(x+1)(2x+1)^2=0

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

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 (0,0) , (-1,0) and $\left ( \frac{-1}{2}, \sqrt{\frac{1}{48}}\right )$ . We need to determine if the critical points are maximums or minimums using D(x,y) 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)=(2y)(24y)-(2x+1)^2=48y^2-(4x^2+4x+1)

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

At (0,0) ,

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

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

At (-1,0) ,

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

D(x,y)=-9<0

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

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

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

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

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

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

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Example Question #1 : Relative Minimums And Maximums

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

Possible Answers:

(1.429,-0.757) is a saddle point.

(1.429,-0.757) is a relative minimum.

(1.429,-0.757) is a relative maximum.

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

Correct answer:

(1.429,-0.757) is a saddle point.

Explanation:

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

to classify the critical points.

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

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

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

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

The first order partial derivatives are

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

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

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

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

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

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

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

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

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

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

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

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

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

36y^3-12y^2-20y-52=0

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 (1.429,-0.757) . We need to determine if the critical point is a maximum or minimum using D(x,y) 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)=(2+6y)(-4)-(6x)^2

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

At (1.429,-0.757) ,

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

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

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Example Question #2 : Relative Minimums And Maximums

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

Possible Answers:

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

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

(0,0) and (-1.6,-1.786) are relative maxima

(0,0) and (-1.6,-1.786) are saddle points

Correct answer:

(0,0) and (-1.6,-1.786) are saddle points

Explanation:

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

to classify the critical points.

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

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

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

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

The first order partial derivatives are

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

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

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

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

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

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

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

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

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

The real values of are x=0 and x=-1.6

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 (0,0) and (-1.6,-1.786) . We need to determine if the critical points are maxima or minima using D(x,y) 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)=(12x-4y)(12y^2)-(-4x-7)^2

At (0,0) ,

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

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

At (-1.6,-1.786) ,

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

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

Since D(x,y)<0 , (-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 f(x,y)=x^4+y^4+3xy .

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

to classify the critical points.

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

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

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

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

The first order partial derivatives are

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

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

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

x=0 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=0 , $x=-\sqrt3/2$ 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 (0,0) , $(-\sqrt3/2,\sqrt3/2)$ and $(\sqrt3/2,-\sqrt3/2)$ . We need to determine if the critical points are maxima or minima using D(x,y) 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)=(12x^2)(12y^2)-(3)^2=144x^2y^2-9

At (0,0) ,

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

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

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Example Question #1 : Relative Minimums And Maximums

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

Possible Answers:

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

(0.0617,-0.2495) is a relative maxima

(0.0617,-0.2495) is a saddle point

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

Correct answer:

(0.0617,-0.2495) is a saddle point

Explanation:

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

to classify the critical points.

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

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

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

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

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

The first order partial derivatives are

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

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

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

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

y=-cos(x)/4

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

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 y=-cos(x)/4 (found by rearranging the first derivative)

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

There is a critical points at (0.0617,-0.2495) . We need to determine if the critical point is a maximum or minimum using D(x,y) 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)=-sin(x)cos(y)-4^2

At (0.0617,-0.2495) ,

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

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

Report an Error

Example Question #9 : Relative Minimums And Maximums

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

Possible Answers:

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

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

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

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

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

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

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

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

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

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

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

Correct answer:

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

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

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

Explanation:

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

to classify the critical points.

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

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

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

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

The first order partial derivatives are

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

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

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

f_x(x,y)=2sin(x)cos(x)=0

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

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

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

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

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

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

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

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

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

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

We need to determine if the critical point is a maximum or minimum using D(x,y) 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)=(2cos^2x-2sin^2x)(2sin^2y-2cos^2y)-0^2

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

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

D(0,0)=(2cos^2(0)-2sin^2(0))(2sin^2(0)-2cos^2(0))-0^2

=(2(1)-0)(2(0)-2(1))=(2)(-2)=-4<0

Saddle point

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

=(2(1)-2(0))(2(1)-2(0))=(2)(2)=4>0

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

minimum

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

=(2(1)-0)(2(0)-2(-1))=(2)(2)=4>0

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

minimum

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

=(2(1)-2(0))(2(-1)-2(0))=(2)(-2)=-4<0

Saddle point

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

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

=(2(0)-2(1))(2(0)-2(1))=(-2)(-2)=4>0

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

maximum

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

=(2(0)-2(1))(2(1)-2(0))=(-2)(2)=-4<0

saddle point

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

=(2(0)-2(1))(2(0)-2(-1))=(-2)(2)=-4<0

saddle point

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

=(2(0)-2(1))(2(-1)-2(0))=(-2)(-2)=4>0

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

maximum

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

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

=(2(-1)-0)(2(0)-2(1))=(-2)(-2)=4>0

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

maximum

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

=(2(-1)-2(0))(2(1)-2(0))=(-2)(2)=-4<0

saddle point

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

=(2(-1)-0)(2(0)-2(-1))=(-2)(2)=-4<0

saddle point

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

=(2(-1)-2(0))(2(-1)-2(0))=(-2)(-2)=4>0

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

maximum

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

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

=(2(0)-2(-1))(2(0)-2(1))=(2)(-2)=-4<0

saddle point

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

=(2(0)-2(-1))(2(1)-2(0))=(2)(2)=4>0

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

minimum

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

=(2(0)-2(-1))(2(0)-2(-1))=(2)(2)=4>0

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

minimum

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

=(2(0)-2(-1))(2(-1)-2(0))=(2)(-2)=-4<0

saddle point

Report an Error

Example Question #4 : Relative Minimums And Maximums

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

Possible Answers:

(0,0) is a relative minimum

(1,1) is a relative minimum

(1,1) is a relative maximum

(0,0) is a relative maximum

Correct answer:

(0,0) is a relative minimum

Explanation:

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

to classify the critical points.

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

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

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

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

The first order partial derivatives are

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

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

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 (0,0) . We need to determine if the critical point is a maximum or minimum using D(x,y) 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)=2(2)-0^2=4

At (0,0) ,

D(x,y)=4>0

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

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

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Calculus 3 : Relative Minimums and Maximums

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