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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 graphing device as in Example 4 (or Newton’s method or solve numerically using a calculator or computer) to find the critical points of f correct to three decimal places. Then classify the critical points and find the highest or lowest points on the graph, if any.

28. f(x, y) = y6 − 2y4+ x2y2 + y

To determine

To find: The critical points of the function f(x,y)=y62y4+x2y2+y and then the highest or lowest point of the function using the graph.

Explanation

Result used:

Second Derivative Test:

“Suppose the second partial derivatives of f are continuous on a disk with center (a,b) , and suppose that fx(a,b)=0 and fy(a,b)=0 (that is (a,b) is a critical point of f).

Let D=D(a,b)=fxx(a,b)fyy(a,b)[fxy(a,b)]2

(a) If D>0 and fxx(a,b)>0 , then f(a,b) is a local minimum.

(b) If D>0 and fxx(a,b)<0 , then f(a,b) is a local maximum.

(c) If D<0 , then f(a,b) is not a local maximum or minimum and it is called a saddle point”.

Given:

The function is, f(x,y)=y62y4+x2y2+y .

Calculation:

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

fx=x[y62y4+x2y2+y]=00+2x0+0=2x

Thus, fx=2x . (1)

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

fy=y[y62y4+x2y2+y]=6y52(4y3)+02y+1=6y58y32y+1

Thus, fy=6y58y32y+1 . (2)

Solve the equations (1) and (2) and find the values of x and y.

From the equation (1),

2x=0x=0

From equation (2),

6y58y32y+1=0

Find the value of y by using the graph of 6y58y32y+1=0 .

Use online graphing calculator and draw the graph of 6y58y32y+1=0 as shown below in Figure 1.

From Figure 1 it can be observed that the values of y are 1.273,0.347and1.211 .

Therefore, the critical points of the function f(x,y)=y62y4+x2y2+y are (0,1.273),(0,0.347)and(0,1.211) .

Obtain the second derivatives as follows.

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

2fx2=x[2x]=2

Hence, 2fx2=2 .

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

2fy2=y[6y58y32y+1]=6(5y4)8(3y2)2(1)+0=30y424y22

Hence, 2fy2=30y424y22 .

Take the partial derivative of the equation (1) with respect to y and obtain fxy .

2fxy=y[2x]=0

Hence, 2fxy=0

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Chapter 14 Solutions

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