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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 ) = 3 x 2 4 y 2 6 x 2 y + 9

To determine

To calculate: The relative extrema and saddle points of the function

f(x,y)=3x24y26x2y+9

Explanation

Given Information:

The function with two variables is given as f(x,y)=3x24y26x2y+9

Formula used:

To find the relative extrema and saddle points of the function, follow the following procedure:

Step 1: Write primary equation for the function to be minimize.

Step 2: Find first partial derivative of primary equation and equate to zero to find critical point.

Step 3: fx(x,y)=0     &    fy(x,y)=0

Find double partial derivative with respect to all variables separately at different critical points.

Suppose (a,b) is critical point, then find double partial derivative like:

fxx(a,b),fyy(a,b) and fxy(a,b) & fyx(a,b) then to test the function f(x,y) is minimum, maximum and having saddle point consider the secondary equation:-

d=fxx(a,b).fyy(a,b)[fxy(a,b)]2

i) If d>0 and fxx(a,b)>0 then function f(x,y) has a relative maximum at (a,b),

ii) If d>0 and fxx(a,b)<0, then function f(x,y) has a relative maximum at (a,1),

iii) If d<0 and [(a,b),f(x,y)] is a saddle point.

iv) The test gives no information when d=0.

Calculation:

Considering the given equation

f(x,y)=3x24y26x2y+9

Differentiating given equation with respect to x keeping y-as constant

fx(x,y)=xf(x,y)fx(x,y)=x[3x24y26x2y+9]fx(x,y)=6x6fx(x,y)=6(x+1)

Equating fx(x,y)=6(x+1) to zero to for critical point

fx(x,y)=06(x+1)=0x=1

Differentiating given equation with respect to y keeping x-as constant

fy(x,y)=yf(x,y)</

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