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?
Best Answer
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.