1. ## 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.

2. 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.

3. ## OK here is what you do

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

4. 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!

5. ## Haha

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