# Optimization Problems

Code:
A 51 meter length of wire is cut into two parts.
The first part is fashioned into a rectangle that is twice as long as it is wide.
The second part is fashioned into a square.
How much of the originial wire is used for each shape
if the shapes' combined area is a minimum?
Use the second derivative test(S.D.T.) to determine the minimum.
Couple of questions/comments on this problem. Why the SDT? Out of all the examples in the bookn either max or min optimization problems don't do the SDT. They do the first derivative test to determine critical values to solve the problem. Second question is my primary and secondary equations set up right? This the primary equation:
$$A = xy + x^2$$
Secondary:
$$51 = 2(2x + y) + 4x$$

Why are you assuming the rectangle has a short side equal to a side of the square? That's not necessarily the case.

So I need a third variable? I'm starting to understand this less and less.

Tom Mattson
Staff Emeritus
Gold Member
No you don't need a third variable, because the length of the wire was given. You've made two mistakes here.

1.) The dimensions of the rectangle aren't 2x by y, they're 2x by x. They told you that one dimension is twice the other.

Then there's the mistake that daveb pointed out.

2.) You're assuming that each side of the square has the same length as the width of the rectangle. That isn't necessarily the case, so call it something else (like y).

HallsofIvy