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The problem I am working on is:

Find the maximum and minimum values of the function:

$f(x,y) = -3x^2 - 14xy - 3y^2 -8$ on the disk: $x^2 + y^2 \leq 4$

The $-14xy$ term is severely throwing me for a loop. It's complicating finding the critical points on the boundary and for the function that lies on the disk. I'm not sure how to approach this problem because of that.

If anyone could walk me through this one, I would greatly appreciate it. I don't have notes that I can refer to for dealing with these types of problems.

Kommander Kitten
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  • Are you able to use lagrange or only regular methods? – Quality Apr 04 '15 at 21:36
  • I think the purpose of the problem is to use Lagrange. But I'm not very comfortable with it. I need to learn it though, so Lagrange would be best. Practice makes perfect! – Kommander Kitten Apr 04 '15 at 21:38
  • Okay I will formulate something, but you said you atleast understand how to do these questions if it was just the interior? – Quality Apr 04 '15 at 22:04
  • Yes. Taking the partial derivatives and finding the critical points. The issue here is the $-14xy$ term that complicates things for me. – Kommander Kitten Apr 04 '15 at 22:09

1 Answers1

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Okay we are looking to find the maximum and minimum values of the equation given by $$f(x,y)=-3x^2-14xy-3y^2-8$$ on the closed disk $x^2+y^2 \le 4$

First as you said we must check the interior. That is

$f_x=0 ; -6x+14y=0$

$f_y=0; -6y-14x=0$

But as you said the issues are on the boundary. In which case we no longer have $x^2+y^2 \le 4$, we have $x^2+y^2=4$. Now we can use lagrange

so $\nabla f_x=\lambda \nabla g_x$ and $\nabla f_x=\lambda \nabla g_y $ and $x^2+y^2=4$

For for $\nabla f_x= -6x-14y= \lambda (2x)$

$\rightarrow -3x-7y=\lambda x$

and $\nabla f_y= -6y-14= \lambda (2y)$

$\rightarrow -3y-7x=\lambda y$

Multiply by y and x respectively to get

$-3xy-7y^2=\lambda xy$

$-3xy-7x^2=\lambda xy$

equate , and solve $-7x^2=-7y^2 \rightarrow x=y$ or $x=-y$

if $x=y$ then $x^2+x^2=4$ that is $2x^2=4 x^2=2 x=+-\sqrt{2}$ , plug into the f to check value

if $x=-y$ then you get the same values for $x$, plug into f to check

And doing so you will see you have the max value of $f(x,y)=8$ at $(x,y)=(-\sqrt{2},\sqrt{2})$ and also at ($x,y)=(\sqrt{2},-\sqrt{2})$ That is when x=-y

and for minimum, when you tried x=y you would get $f(x,y)=-48$ at $(\sqrt{2},\sqrt{2})$ and at $(-\sqrt{2},-\sqrt{2})$

Quality
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