Wire Optimization

Finally, after waiting for a good question, it finally came: an optimization question. This one is from Yahoo Answers, and believe me, it's very good.

A 6 meter long wire is cut into 12 pieces. From these pieces, eight have the same length and the other four also have an equal length. These pieces form a frame of the box. How long should each piece for the box to have a maximum volume?

First, we should define ourselves some variables, otherwise we'd be lost.

Let x be the length of one of the 8 pieces. That means the total length of the 8 equal pieces is 8x. That means the other four pieces have a total length of 6 - 8x, and 3/2 - 2x meters each.

Since these wires form a box, the sides of the box have lengths of x, x, and 3/2 - 2x. That makes the volume of the box x * x * (3/2 - 2x), or 3x²/2 - 2x³.

To find the maximum value, we need to differentiate the volume function and set it to zero. Then we need to find the values that we get from solving the equation, plug them in the second derivative, see which one is a maximum and say "Problem Solved".

So:
V = 3x²/2 - 2x³
V' = 3x - 6x²
0 = 3x - 6x²
3x(1 - 2x) = 0
x = 0, 1/2

Now, let's find the second derivative and see when it's negative, so we get a maximum:
V'' = 3 - 12x
V''(0) = 3 - 0 = 3 > 0 - minimum
V''(1/2) = 3 - 12 * 1/2 = 3 - 6 = -3 < x =" 1/2" 1 =" 1/2">3 = 1/8 m³

Problem solved.
Nadav

nadavs