   Chapter 7.5, Problem 26E ### 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 − 3 x 2 + 6 y 2 + 3 x + 12 y + 7

To determine

To calculate: The critical points of the function f(x,y)=x3+y33x2+6y2+3x+12y+7 and the relative extrema and the saddle points

Explanation

Given Information:

The function with two variables is f(x,y)=x3+y33x2+6y2+3x+12y+7

Formula used:

Partial derivatives of the function with respect to x as well as y,

fx=f(x,y)xfy=f(x,y)y

Double partial derivatives of the function

fxx=2f(x,y)x2fyy=2f(x,y)y2fxy=2f(x,y)xy

At the critical point if the value of D=fxx(a,b)fyy(a,b)[fxy(a,b)]2,

D>0 and r<0, then declare the point as a maxima.

D>0 and r>0, then declare the point as a minima.

D<0, then this critical point will not be an extremum point.

D=0, then say double partial derivative fails and need further investigation on this point

Calculation:

Consider the given equation,

f(x,y)=x3+y33x2+6y2+3x+12y+7

Now, compute the first partial derivatives and find the critical points while equating to zero

fx=0

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