# Homework Help: Max problems

1. Apr 13, 2010

### apiwowar

the top and bottom margins of a poster are 2cm and the side margins are each 4cm. If the area of printed material on the poster is fixed at 382 square centimeters, find the dimensions of the poster with the smallest area.

so the area l*w = 382

and the perimiter = 2(l-4)+2(w-8)

but if i try to solve for l i get a negative number. can someone help me out?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 14, 2010

### HallsofIvy

You are setting it up wrong. You say " l*w = 382" but 382 is the area of the printed material so the actual dimensions are l+4 and w+8. Also the perimeter is irrelevant. You are asked to find l and w such that the total area, (l+ 4)(w+ 8)= lw+ 4w+ 8l+ 32, is a minimum.

From lw= 382, w= 382/l and that becomes 382+ 4(382)/l+ 8l+ 32= 6112/l+ 8l+ 412. Minimize that.

Finally, remember that the answer, the dimensions of the poster, are l+ 4 and w+ 8, not l and w.