Absolute Max/Min of f(x,y): Find Solution

[camino]

Homework Statement

Find the absolute maximum and absolute minimum of f(x,y)=12xy-x^2 y-2xy^2 on the region bounded by x=1, y=1, and y=4/x

Homework Equations

The Attempt at a Solution

For x=1
f(1,y)= 12y-y-2y^2
= 11y-2y^2
f'(1,y)= -4y+11=0
= -4y=-11
= y=11/4

For y=1
f(x,1)= 12x-x^2-2x
= 10x-x^2
f'(x,1)= -2x+10=0
= -2x=-10
= x=5

For y=4/x
I have no idea.

Am I even doing these right? I am confused. Any work/help would be greatly appreciated!

 

[Chaos2009]

I haven't worked much with functions like this with two variables, but usually when you are trying to find a maximum or minimum, especially in calculus, you use the derivative. You look at the critical points of your functions which are when the derivative equals zero, the derivative is undefined, and the endpoints. Those are the places where maximums and minimums could occur. To find out which ones are the absolute max or min, you have to go back and plug the values in and see.

 

[Dick]

Did you draw sketch of your region? Do you know what it looks like? So far you are only looking for points on two boundaries of the region. But the points you have found may be outside of the region. Are they? To handle the y=4/x boundary just substitute y=4/x into f. Finally you need to think about possible extrema inside the region. Set the partial derivative of f with respect to x and y equal to zero and solve simultaneously.

 

[camino]

f(x,4/x)=12x(4/x)-x^2(4/x)-2x(4/x)^2
that is substituting x/4 into f, and it makes for a messy solving method with I don't think is right. There has to be an easier way to do that part that I'm just not seeing.

fx(x,y)=12y-2xy-2y^2
fy(x,y)=12x-x^2-4xy
That is solving for the interior, and it also is messy.

It seems to me that there should be a simple way to achieve these answers, but I just can not figure it out. This is due tomorrow and I just can't get it..