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I have this problem:

Find the critical points and say whether they are maxima, minima or saddle points

$$f(x,y)=x^2y(2−x−y)$$

My answer:

$f_x = xy (4-3 x-2 y) $

$f_y = -x^2 (-2+x+2 y)$ then

$xy (4-3 x-2 y)=0 , -x^2 (-2+x+2 y)=0$ the solutions are:

1)x=0 and $y \in{} \mathbb{R}$

2)x=1 and y=1/2

3)x=2 and y=0

I have three critical points,

$f_{xx} = -2 y (-2+3 x+y) $

$f_{yy} = -2 x^2 $

$f_{xy} = x (4-3 x-4 y) $, using

$D=f_{xx}f_{yy}-(f_{xy})^2 = -16 x^2+24 x^3-9 x^4+24 x^2 y-12 x^3 y-12 x^2 y^2$

x=1 , y=1/2 then D=0 but $f_{xx}=-2y (-2+3 x+y)= -2/3$ concluding (1,1/2) is a maxima point

x=2 , y=0 then D=-16 we say that (2,0) is a saddle points

x=0 , $y \in{} \mathbb{R}$ then D=0 the point (x,a) with $a \in{} \mathbb{R}$ is a saddle points because for example if x=0 , a=-8 $fxx=-160$ (maxima) but if x=0 , a=1 then $f_{xx}=2$ (minima)

Is my exercise correct?

Pedro
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user63192
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    I agree with all of your work, except classifying the three points. The "local Max" is correct. the other two, you need to revisit. See my posting at http://math.stackexchange.com/questions/369945/maximum-and-minimum-absolute-of-a-function-x-y/369956#369956 for two approaches. You need to look at g(x, y) and $f_{xx}$. – Amzoti May 29 '13 at 03:57

1 Answers1

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Look at the Hessian matrix and consider its eigenvalues. If the determinant is negative, then you get a saddle point. If the determinant is zero, the test is inconclusive; if the determinant is positive and $f_{xx} > 0$, then you get a minima; and if the determinant is positive and $f_{xx} < 0$ , then you have a maxima. Now, can you see whether you are correct?

Hessian matrix is just the matrix

\begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \\ \end{pmatrix}

Pedro
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Vishal Gupta
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