1
$\begingroup$

Here's the problem

For each of the following functions, find the maximum and minimum values of the function on the rectanglar region: $−3≤x≤3,−4≤y≤4$. Do this by looking at level curves and gradiants.

I was able to solve the first two function, however I cannot find the right answer to the last function.

This function(click on the link to see)

My attempt:

I first found

$F_x=32x$

$F_y=-18y$

Then I equated these to $0$, resulting in $(x,y)=(0,0)$ and found

$F_{xx}=32$

$F_{yy}=-18$

$F_{xy}=0=F_{yx}$

$D=32*(-18)-(0)^2=-576$

Thus, $D<0$ , thus we have a saddler point. So I'm very lost. I tried inputting $(0,0)$, but it's wrong.

I tried also in putting value $(-3,-4);(-3,4);(3,-4);(3,4)$ which all results into zero from the formula.

Can someone please explain what I am doing wrong?

$\endgroup$
  • $\begingroup$ For a maximum you want $4^2x^2$ as positive as possible and $-3^2y^2$ as positive as possible. For a minimum you want $4^2x^2$ as negative as possible and $-3^2y^2$ as negative as possible. $\endgroup$ – Henry Dec 3 '16 at 1:33
  • $\begingroup$ So for a maximum, I would have to have (x,y)=(+/-4,0) ; and for a minimum (x,y)=(0,+/-4) if I understand well $\endgroup$ – lola Dec 3 '16 at 1:36
  • $\begingroup$ local minimizer $ \nabla f(x,y)=0$, for global minimizer you have to check on those points which arent inner points, namly the boundarys, too. A general way would be to paramize the boundarys (4 straight lines) and compare every single min and max there, if f would no so obviously easy. $\endgroup$ – user160069 Dec 3 '16 at 1:39
  • $\begingroup$ You also have to check the boundary conditions, that is, the values that $f$, that is, the values that f has in all the border of the rectangle $\endgroup$ – GLay Dec 3 '16 at 1:43
  • $\begingroup$ @user160069 I'm not sure I completely understand. We take the gradient -> ∇f=32x-18y and then we should input numbers like -4,4 for x, and -5,5 for y? in this case we can see from the gradient that the maximum is (x,y)=(4,-5) ? Are we supposed to only look at the gradient in case of saddler points? $\endgroup$ – lola Dec 3 '16 at 1:45
0
$\begingroup$

What you have done is found that there is a local critical point at $(0,0)$, and that it is a saddle point.   The function is a hyperbolic parabaloid.   So indeed it is neither a maximum nor a minimum.

What you need to do is examine the boundaries of the interval; which is a rectangle, $[-3;3]{\times}[-4;4]$.   What you have done wrong is only look at the four corners of the rectangle; rather than along the sides.

So, when $x=-3$ the maximum of $f(-3,y)= 16(-3)^2-9 y^2$ is located where ?

...and so forth.

$\endgroup$
  • $\begingroup$ I think I understand now. For example in this case the maximum would be at (-3,0) , but won't that give me two maximum? (-3,0) and (3,0)? $\endgroup$ – lola Dec 3 '16 at 2:24
  • $\begingroup$ Yes, @lola , it is possible to have multiple maxima and minima in the interval, and in this case there are. Look at all four extremes of the interval (aka sides of the rectangle). $\endgroup$ – Graham Kemp Dec 3 '16 at 2:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.