Question

Find the dimensions of the rectangle of maximum area that can be formed from a 330-in. piece of wire.

Answer 1

make a square

Answer 2

Okay, start out by identifying your variable, so let w = width of the rectangle. Then, from the fact that it is a rectangle, you know the perimeter must add to 330, so you can get your other side in terms of w:

330 = 2w + 2l
l = 165 - w

Then,

A = w(165 - w)
= 165w - w^2

Then, take the derivative to find your critical point. Sinve we can see that this parabola opens downwards, you don't need to worry about using the second derivative to make sure it's a maximum or minimum.

A' = 165 - 2w
0 = 165 - 2w
w = 165/2

Which is exactly a quarter of your total length, so you can see you have a square of side length (165/2).

Answer 3

Area is \(W\times L\) and since you cannot tell \(W\) from \(L\), i.e. it is completely symmetric in \(W\) and \(L\) it must be the biggest when \(W=L\)