Answer to Question #138349 in Calculus for Promise Omiponle

Question

Consider the function
f(x,y)=(2x−x^2)(8y−y^2).
Find and classify all critical points of the function.
First list out all the first and second partial deriatives.
There are several critical points to be listed. List them lexicograhically, that is in ascending order by x-coordinates, and for equal x-coordinates in ascending order by y-coordinates (e.g., (1,1), (2, -1), (2, 3) is a correct order)

Answer 1

Given that,

$$f(x,y)=(2x-x^2)(8y-y^2)$$

Clearly $f\in \mathscr{C}^{\infty}$ , thus Critical points are given by

$$\nabla f(x,y)=((2-2x)(8y-y^2),(2x-x^2)(8-2y))=(0,0)$$

Then we get,

$$(2-2x)(8y-y^2)=0\\ (2x-x^2)(8-2y)=0$$

After solving the above equation we get the critical points are ,

$$(0,0),(0,8),(1,4),(2,0),(2,8)$$

Now, calculate the Hessian matrix

$$Hf(x,y)=\begin{bmatrix} f_{xx}& f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix}$$

Thus in this case we get

$$Hf(x,y)=\begin{bmatrix} -2(8y-y^2)&4(1-x)(4-y)\\ 4(1-x)(4-y)&-2(2x-x^2) \end{bmatrix}$$

Thus, $Hf(0,0),Hf(0,8),Hf(2,0),Hf(2,8)$ are indefinite as eigen values are $\pm 4$ , thus (0,0) ,(0,8),(2,8) are a saddle point.

Similarly, $Hf(1,4)$ is negative definite as eigen values are -2,-32,thus (1,4) is point of minima.