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



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





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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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