   Chapter 7.5, Problem 25E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Analyzing a Function In Exercises 25–30, find the critical points, relative extrema, and saddle points of the function. List the critical points for which the Second-Partials Test fails. f ( x , y ) = x 3 + y 3

To determine

To calculate: The critical points, relative extrema and saddle points of the function f(x,y)=x3+y3. Also determine the critical points at which the second partial test fails.

Explanation

Given Information:

The provided function is f(x,y)=x3+y3.

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+y3

The first partial derivatives of the function f are,

fx(x,y)=x[x3+y3]=3x31+0=3x2

And

fy(x,y)=y[x3+y3]=0+3y31=3y2

Equate both partial derivatives to zero,

3x2=0x=0

And

3y2=0y=0

So, the critical point is (0,0)

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