# Math Help - Quadratics Problem

A 30 cm piece of wire is cut into two. One piece is bent into the shape of a square, the other piece into the shape of a rectangle with a length-to-width ratio of 2:1. What are the lengths of the two pieces if the sum of the areas of the square and rectangle is a minimum?

This is what I did so far:

Let x represent the width of the rectangle
" y " square
" z " sum of the areas

30=2x+2x+x+x+4y
=6x+4y
7.5=y+1.5x
y=7.5-1.5x

z=y^2+2x^2
=(7.5-1.5x)^2+2x^2
=4.25x^2-22.5x+56.25

but then when I try finding the vertex, by completing the square, I don't get an integer. I'm not sure what to do.

2. Originally Posted by DemonX01
A 30 cm piece of wire is cut into two. One piece is bent into the shape of a square, the other piece into the shape of a rectangle with a length-to-width ratio of 2:1. What are the lengths of the two pieces if the sum of the areas of the square and rectangle is a minimum?

This is what I did so far:

Let x represent the width of the rectangle
" y " square
" z " sum of the areas

30=2x+2x+x+x+4y
=6x+4y
7.5=y+1.5x
y=7.5-1.5x

z=y^2+2x^2
=(7.5-1.5x)^2+2x^2
=4.25x^2-22.5x+56.25

but then when I try finding the vertex, by completing the square, I don't get an integer. I'm not sure what to do.
Providing your calculations to this point are correct, you need to differentiate, so this should really be in the calculus forum, but oh well.

$\frac{dy}{dx} = 9.5x - 22.5$

To find the coordinate of the stationary point, you equate $\frac{dy}{dx} = 0$

So, $9.5x - 22.5 = 0$

$x = 2.368421053$

Plug back in the equation to find z. I've got to go now, so I can't do it, but that's how I'd do it.

3. I've already considered diffrentiating, but it doesn't produce an interger, so something in my calculations is probably off. I just need to know what.