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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 ) = 5 x 2 + 4 x y y 2 + 16 x + 10

To determine

To calculate: The relative extreme and saddle point of the function f(x,y)=5x2+4xyy2+16x+10.

Explanation

Given Information:

The provided function f(x,y)=5x2+4xyy2+16x+10

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)=5x2+4xyy2+16x+10

The first partial derivatives of the function f with respect to x and y are,

fx(x,y)=10x+4y+16fy(x,y)=4x2y

Equate fx(x,y)=0,

10x+4y+16=05x+2y+8=0

Equate fy(x,y)=0,

4x2y=0y=2x

Substitute y=2x in the equation 5x+2y+8=0,

5x+2(2x)+8=0x+8=0x=8

Substitute x=8 in the equation y=2x,

y=2(8)=16

So, the critical point is (8,16)

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