Question #138349
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)
1
Expert's answer
2020-10-22T18:12:03-0400

Given that,


f(x,y)=(2xx2)(8yy2)f(x,y)=(2x-x^2)(8y-y^2)

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

f(x,y)=((22x)(8yy2),(2xx2)(82y))=(0,0)\nabla f(x,y)=((2-2x)(8y-y^2),(2x-x^2)(8-2y))=(0,0)

Then we get,


(22x)(8yy2)=0(2xx2)(82y)=0(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)(0,0),(0,8),(1,4),(2,0),(2,8)

Now, calculate the Hessian matrix


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

Thus in this case we get


Hf(x,y)=[2(8yy2)4(1x)(4y)4(1x)(4y)2(2xx2)]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)Hf(0,0),Hf(0,8),Hf(2,0),Hf(2,8) are indefinite as eigen values are ±4\pm 4 , thus (0,0) ,(0,8),(2,8) are a saddle point.

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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS