# Homework Help: Optimization problem

1. Jul 8, 2011

### QuarkCharmer

1. The problem statement, all variables and given/known data
I took a test today on integration, curve sketching, and optimization. I am pretty sure that I got a 100 on it due to all the help here on PF with indef. integration and all of the helpful u-sub advice I have received. Anyway, there was 5 optimization word problems, and only 4 were counted towards the grade. I did all 5 of them and I know 4 were correct, but the 5th one I could not figure out how to set up. It was something like this:

Someone is trying to make a rectangular fence, and they want to use a fancier fence material for the front side of the house. Material for the back and sides cost 2 dollars per foot, and material for the front costs 3 dollars per foot. This person only has 120 dollars to spend. What is the largest possible area they can cover?

2. Relevant equations

3. The attempt at a solution

I claimed that the sides were L in length, and therefor the cost of the sides would be 2(2L). Then I said that the length of the front would be W, and the cost of the front would be 3W. Now I claimed the back wall would be 2W (because it's the same length as the front, only with the cheap material). So that the cost of the whole thing is given by:
$4L+5W=120$

I am not sure how to pull another equation out of this one so I can substitute for W or L or something. There was no specification like "The area is something" that I could use for that.

Edit: Wrong forum section sorry, it's close enough to pre-calculus algebra though.

Last edited: Jul 8, 2011
2. Jul 8, 2011

### Staff: Mentor

The expression you want to maximize is the area, or LW, subject to the constraint that 4L + 5W = 120. Solve this equation for one of the variables, and then substitute it into your area expression to get area as a quadratic function of a single variable. The graph of this function will be a parabola. It's a good bet that the parabola will open downward...

3. Jul 8, 2011

### SammyS

Staff Emeritus
The area A is A = L · W .

Solve the cost equation for L or W & plug into the Area equation. Then find the maximum for A.

(Since A will be quadratic in L or W, this is a problem which could be done in College Algebra.)