MHB Local Min Problem: Solve 2a>0 & 4a2-1 > 0?

[Yankel]

Hello all,

I have this tricky question, I think I got the idea, just wish to confirm.

If the function

\[z=x\cdot ln(1+y)+a(x^{2}+y^{2})\]

has a local minimum at (0,0), then: (choose correct answer)

1) a<-0.5
2) a>0
3) a>0.5
4) -0.5<a<0.5
5) a>0.5 or a<-0.5

What I did, is calculate the partial derivatives by x and y. Then I calculated the second order partial derivatives to get D, the hessian matrix. I then put (0,0) in the hessian matrix, and got this condition:

\[4a^{2}-1>0\]

I solved it to get

a>0.5 or a<-0.5

but, I also looked just on the second derivative by x and x. It was equal to 2a. I know it must be bigger than 0 (minimum), thus a>0

so from both conditions I conclude that (3) is the correct answer.

Am I right ?

Thank you !

 

[Sudharaka]

Hello all,

I have this tricky question, I think I got the idea, just wish to confirm.

If the function

\[z=x\cdot ln(1+y)+a(x^{2}+y^{2})\]

has a local minimum at (0,0), then: (choose correct answer)

1) a<-0.5
2) a>0
3) a>0.5
4) -0.5<a<0.5
5) a>0.5 or a<-0.5

What I did, is calculate the partial derivatives by x and y. Then I calculated the second order partial derivatives to get D, the hessian matrix. I then put (0,0) in the hessian matrix, and got this condition:

\[4a^{2}-1>0\]

I solved it to get

a>0.5 or a<-0.5

but, I also looked just on the second derivative by x and x. It was equal to 2a. I know it must be bigger than 0 (minimum), thus a>0

so from both conditions I conclude that (3) is the correct answer.

Am I right ?

Thank you !

Hi Yankel, :)

If in the problem it is given that $a>0$, then your answer is correct. (Yes)