# Maximizing a function.

• April 8th 2008, 02:15 PM
zsig013
Maximizing a function.
Story problems always confuse me. Can someone point me in the right direction? Thanks!

There are 544 yards of fencing available to enclose a rectangular field. How should this fencing be used so that the enclosed area is as large as possible? The area is maximized at length = ____ yards and width = ____ yards.
• April 8th 2008, 02:49 PM
TrevorP
First create an equation for the area in terms of width and height.
Find an equation of perimeter (2x width + 2x height).

Isolate for one variable...it should be a parabola. Then if you are in calculus take the derivative and solve for 0. If not find the midpoint with -b/2a.
• April 8th 2008, 02:50 PM
Mathstud28
OK here is what you do
Quote:

Originally Posted by zsig013
Story problems always confuse me. Can someone point me in the right direction? Thanks!

There are 544 yards of fencing available to enclose a rectangular field. How should this fencing be used so that the enclosed area is as large as possible? The area is maximized at length = ____ yards and width = ____ yards.

since it is a rectangle you know that $544=2x+2y,272=x+y$ and you know that Areas which is what you are trying to maximize is $A=xy$...using the substitution $y\cdot(272-y)=272y-y^2=A$ we get our function we want to maximise diffrentiating we get $A'=272-2y$...which equals zero when $y=136$...before we yell hooray we need to check if it is a max or min so by diffentiating again we get $A''=-2$ so we know that $y=136$ is a max now subbing that back into your $272=x+y$ you can get your X value
• April 8th 2008, 02:52 PM
zsig013
Okay, I took the derivative and get x=136. Making it a square (136*4=544). So It was simple as that Length and width both are 136? EDIT: Ahh Mathstud you're too quick. Thanks a lot!
• April 8th 2008, 03:16 PM
Mathstud28
Haha
Quote:

Originally Posted by zsig013
Okay, I took the derivative and get x=136. Making it a square (136*4=544). So It was simple as that Length and width both are 136? EDIT: Ahh Mathstud you're too quick. Thanks a lot!

Anytime thats what I come on for