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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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Applying the Second-Partials Test In Exercises 1-18, find the relative extrema and saddle points of the function. See Examples

f ( x , y ) = x 3 + 4 x y 2 y 2 + 1

To determine

To calculate: The relative extreme and saddle point of the function f(x,y)=x3+4xy2y2+1.

Explanation

Given Information:

The provided function f(x,y)=x3+4xy2y2+1.

Formula used:

Second partial test to determine relative extrema.

If the function f have continuous second partial derivative in the open region (a,b) for which fx(a,b)=0 and fy(a,b)=0.

Steps to test relative extrema of the function f,

Consider the quantity, d=fxx(a,b)fyy(a,b)[fxy(a,b)]2

1. The function f contains a relative minimum at (a,b) if the quantity d>0 and fxx(a,b)>0.

2. The function f contains a relative maximum at (a,b) if the quantity d>0 and fxx(a,b)<0.

3. The point (a,b,f(a,b)) is a saddle point if the quantity d<0.

4. If the quantity d=0 then the test gives no information.

In any case, if d>0 then fxx(a,b) and fyy(a,b) must contain same sign. The derivative fxx(a,b) can be replaced by the derivative fyy(a,b) in the first two steps.

Calculation:

Consider the function,

f(x,y)=x3+4xy2y2+1

The first partial derivatives of the function f with respect to x is,

fx(x,y)=x(x3+4xy2y2+1)=3x2+4y

The first partial derivatives of the function f with respect to y is,

fy(x,y)=y(x3+4xy2y2+1)=4x4y

Equate fx(x,y)=0,

3x2+4y=04y=3x2

Equate fy(x,y)=0,

4x4y=0x=y

Substitute x=y in the equation 4y=3x2,

4y=3y2y(3y4)=0

The obtained values are y=0,43.

If y=0 then x=0.

Substitute y=43 in the equation 4y=3x2,

443=3x2169=x243=x

The critical points are (0,0) and (43,43).

The second partial derivatives of the function f are,

fxx(x,y)=6xfyy(x,y)=4fxy(x,y)=4

The values of the second partial derivatives of the function f at critical point (0,0) are,

fxx(0,0)=0fyy(0,0)=4fxy(0,0)=4

The values of the second partial derivatives of the function f at critical point (43,43) are,

fxx(43,43)=8fyy(43,43)=4fxy(43,43)=4

Consider the equation, d

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