a developer wants to enclose a grassy lot that borders a city street for parking. If the developer has 228ft of fencing & does not fence the side along the street, what is the pargest area that can be enclosed?

I would assume you would start off with the area formula where A=LxW, but change it to A=L+2W because of the missing side?

Question

a developer wants to enclose a grassy lot that borders a city street for parking. If the developer has 228ft of fencing & does not fence the side along the street, what is the pargest area that can be enclosed? 

 I would assume you would start off with the area formula where A=LxW, but change it to A=L+2W because of the missing side?

zepdrix · Answer

Your first AREA formula looks good, yes :)
The second thing you wrote down, should be your perimeter.
P=L+2W.

Hmm so we're given an amount of fencing, so perimeter will be our constraint.
And we want to MAXIMIZE the area function.

anonymous · Answer

soo.. 228=L+2w?

zepdrix · Answer

yes good c:

zepdrix · Answer

So the whole idea is, we want to rewrite the AREA function in terms of ONE VARIABLE.

That way we'll be able to differentiate it very easily. We won't have to worry about any freaky prime terms popping out, or applying the product rule :D

So we want to use the information from our Perimeter equation to fill in either the W or L in the area formula.

anonymous · Answer

and, how would I go about doing that?

zepdrix · Answer

$$\large 228=L+2W$$Let's solve this equation for L, then we'll plug that L into the area function. Subtracting 2W from both sides gives us,$$\large L=228-2W$$

Then let's plug it into the area function,$$\large A=L\cdot W \qquad \rightarrow \qquad A=(228-2W)\cdot W$$

anonymous · Answer

okay, so I got A=228W - 2W^2

anonymous · Answer

and then do  I solve for W, and reenter the new values into the equation?

zepdrix · Answer

Sorry was working on another problem c:

So we've found a formula for area involving ONLY W, that's very helpful.
Now we can take the derivative of the Area function, and the look for critical points. The particular critical point that we care about, is the one that is a maximum of the area function. That will maximize area for us.

Can you find the derivative ok? :) It will be with respect to W.

anonymous · Answer

uhhh, no? :/

zepdrix · Answer

$$\large A=228W-2W^2$$Taking the derivative with respect to W, gives us,$$\large A'=228-4W$$Look ok? :)
Lemme make sure I'm not getting ahead of myself, this is a calc 1 problem yes? :D Hopefully I'm not introducing something that you've never seen before.

anonymous · Answer

pre-calc.. I'm confused.. where did the 4W come from? and where did the square root go?

zepdrix · Answer

Oh, hmmmmm :( I'm not sure how to solve optimization problem without using calculus :C Maybe @hartnn does? D':