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Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643

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BuyFindarrow_forward

Multivariable Calculus

8th Edition
James Stewart
ISBN: 9781305266643
Textbook Problem

Use a graph or level curves or both to estimate the local maximum and minimum values and saddle point(s) of the function. Then use calculus to find these values precisely.

24. f ( x , y )   = ( x y ) e x 2     y 2

To determine

To find: The local maximum, local minimum and saddle point using a graph or level curves and use calculus to find these values for the given function f(x,y)=(xy)ex2y2 .

Explanation

Given:

The function is, f(x,y)=(xy)ex2y2 .

Calculation:

Take the partial derivative in the given function with respect to x and obtain fx .

fx=x((xy)ex2y2)=[ex2y2(1)+(xy)ex2y2(2x)]=ex2y22x(xy)ex2y2

Thus, fx=ex2y22x(xy)ex2y2 . (1)

Take the partial derivative in the given function with respect to y and obtain fy .

fy=y((xy)ex2y2)=[ex2y2(1)+(xy)ex2y2(2y)]=ex2y22y(xy)ex2y2

Thus, fy=ex2y22y(xy)ex2y2 (2)

Set the above partial derivatives to 0 and find the values of x and y.

From the equation (1),

ex2y22x(xy)ex2y2=0ex2y2[12x(xy)]=0ex2y2(12x2+2xy)=012x2+2xy=0

From the equation (2),

ex2y22y(xy)ex2y2=0ex2y2[1+2y(xy)]=0ex2y2(1+2xy2y2)=012y2+2xy=0

Subtract the two equations 12x2+2xy=0 and 12y2+2xy=0 .

12x2+2xy(12y2+2xy)=012x2+2xy1+2y22xy=02y2=2x2y=±x

Substitute y=x in the equation 12x2+2xy=0 ,

12x2+2x(x)=012x2+2x2=01=0

This is not possible. So, there is no critical point.

Substitute y=x in the equation 12x2+2xy=0 ,

12x2+2x(x)=012x22x2=01=4x2x=±12

If x=12 then y=12 .

If x=12 then y=12 .

Thus, the critical points are (12,12) and (12,12) .

Obtain the second derivatives as follows.

Take the partial derivative of the equation (1) with respect to x and obtain fxx .

2fx2=x(ex2y22x(xy)ex2y2)=x(ex2y2)x((2x22xy)ex2y2)=[ex2y2(2x)][ex2y2(4x2y)+(2x22xy)ex2y2(2x)]=6xex2y2+2yex2y2+4x3ex2y24x2yex2y2

Hence, 2fx2=6xex2y2+2yex2y2+4x3ex2y24x2yex2y2 .

Take the partial derivative of the equation (2) with respect to y and obtain fyy .

2fy2=y(ex2y22y(xy)ex2y2)=y(ex2y2)y((2xy2y2)ex2y2)=[ex2y2(2y)][ex2y2(2x4y)+(2xy2y2)ex2y2(2y)]=2xex2y2+6yex2y2+4xy2ex2y24y3ex2y2

Hence, 2fy2=2xex2y2+6yex2y2+4xy2ex2y24y3ex2y2

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