# Absolute Max & Absolute Min

1. Mar 5, 2006

### DieCommie

I have been understanding these so far but this one it says is wrong, but I cant figure out why.

"Find the absolute maximum and absolute minimum of the function f(x,y) = xy - 5 y - 25 x + 125 on the region on or above y = x^2 and on or below y = 29."

I get the absolute max to be =148.15 at (-5/3,25/9), which is correct.
I get the absolute min to be =0 at (5,25), which is wrong, but I cant find any more points of interest, as far as I can see 0 at (5,25) is right :grumpy:

If you have any clues/suggestions, I would greatly appreciate it

2. Mar 5, 2006

### Muzza

You have not properly examined what happens to f on the boundary of the region.

3. Mar 5, 2006

### DieCommie

which boundry? y=29 or y=x^2?

for y=x^2 i get the points (5,25) and (-5/3,25/9)
for y=29 i get no points

I have looked over each one a half dozen times, if you know which one the problem is in, id like to know thx.

Last edited: Mar 5, 2006
4. Mar 5, 2006

### Muzza

Well, I know there's at least one point on "the y = 29 boundary" where f is negative. When you consider what happens to f restricted to y = 29, you are checking the endpoints of the interval (which x can range over) and not just the points where the derivative equals 0, right?

5. Mar 5, 2006

### DieCommie

ahhh, because at the endpoints it may be an extrema yet the derivative is non-zero. Got it! Thx alot!

6. Mar 5, 2006

### DieCommie

so if my range is a circle, there are no endpoints to check right?

7. Mar 5, 2006

### HallsofIvy

But the entire circle would be the boundary. Write the boundary in parametric equations, plug those parametric equations into the object function to get a function of one variable and then differentiate that.

In this example, where the boundary was the parabola y= x2 and y= 29, you would need to look for max and min of
f(x, 29)= 29x- 5(29)- 25x+ 125= 4x- 145+ 125= 4x- 20 for the y= 29 part of the boundary. That never has derivative 0 so you look for max and min at its endpoints, $x= \pm \sqrt{29}$.

f(x,x2= x3- 5x2- 25x+ 125 has derivative 3x2- 10x- 25 which is never 0 and so you look for max and min at its endpoints, also $x= \pm \sqrt{29}$.

Last edited by a moderator: Mar 5, 2006
8. Mar 5, 2006

### DieCommie

ugg, sorry, but now Im even more confused on the next one.

"Find the absolute maximum and absolute minimum of the function f(x,y) = 2x^3 + y^4 on the region {(x,y) | x^2 + y^2 =< 16} "

When writing the boundry in parametrics, i use x=cos(theta) y=sin(theta)? THis gives me 1=<16 for my boundry...

9. Mar 6, 2006

### HallsofIvy

You have the wrong parametric equations. The boundary is the circle
x2+ y2= 16 (the interior of the circle, < 16, is not the boundary!). The parametric equations are x= 4 cos($\theta$), y= 4 sin($\theta$).

And you don't put those into the equation for the boundary (which would give you the trivial 16= 16), you put them into the equation for f:
f(x,y)= f($\theta$)= 2(4)3cos3($\theta$)+ 44sin4($\theta$).

10. Mar 6, 2006

### DieCommie

Thx for all your help, I really do appreciate it.

I have been trying and trying this but Im simply unable to do it. I have the derivative of the boundry function in terms of theta, but I do not know how to solve it and find what values of theta give me 0. Im gonna go to the library and try to find a trig. book.