# Find the minimum and maximum of a multi variable function f(x,y)

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.

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?

• 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. – Henry Dec 3 '16 at 1:33
• 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 – lola Dec 3 '16 at 1:36
• 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. – user160069 Dec 3 '16 at 1:39
• 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 – GLay Dec 3 '16 at 1:43
• @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? – 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 ?