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Q. 39

Found in: Page 989


Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

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Short Answer

Extrema: Find the local maxima, local minima, and saddle points of the given functionsf(x, y) = x3+ y3 6x2 + 3y2-4.

Critical point are (0,0) ,(0,-2),(4,0),(4,-2)

See the step by step solution

Step by Step Solution

Step 1. Given 

f(x, y) = x3+ y3 6x2 + 3y2-4.

Step 2. Finding critical point .

Given function is f(x,y)= x3+ y3 6x2 + 3y2-4, which is a polynomial and hence differentiable. The gradient of this function is f(x, y)=fxi+fyj =(3x2-12x)i+(3y2+6y)j At the critical points the gradient of a function vanishes, that is f(x,y)=0. from the above the critical points are given by 3x2-12x =0...(1) and 3y2+6y =0.....(2)from (1) 3x2-12x=0, the roots of which are x=0,4from (1) 3y2+6y=0, the roots of which are y=0,-2Solving these , critical point as (0,0) ,(0,-2),(4,0),(4,-2).

Step 3. Finding maxima and minima .

Now the second order derivatives of the given function are 2fx2=6x-12, 2fy2=6y+6,2fyx=-0.Hence the discriminate of the given function isHf(x,y) =2fx22fy2-(2fyx)2 = (6x-12)(6+6y)-(0)2=36(xy+x-2y-2).At (0,0) , Hf(0,0)=-72<0So the critical point (0,0) is a saddle point .At (0,-2) , Hf(0,-2)=72>0also at (0,-2) 2fx2=-12<0 , 2fy2=-6<0Therefore the critical point is 0,-2 is local maxima. At (4,0) , Hf(4,0)=72>0also at (4,0) 2fx2=12>0 , 2fy2=6>0Therefore the critical point is 4,0 is local minima. At (4,-2) , Hf(4,-2)=-72<0So the critical point (4,-2) is a saddle point.

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