Question

Bob wants to cut a wire that is 60 cm long into two pieces. Then he wants to make each piece into a square. Determine how the wire should be cut so that the total area of the two squares is a small as possible?

Answer 1

The 60 cm long wire should be cut so that you have 2 lengths of 30 cm each.

Explanation:

Let 1 of the cut pieces total length be `x`
Then the other piece is length `60-x`

Let the area for square 1 be `A_1`
Let the area for square 2 be `A_2`
Let the sum of the areas be `A_s`

All sides of a square are of equal length so:

`A_1=(x/4)^2`

`A_2=((60-x)/4)^2`

`A_s=A_1+A_2=(x/4)^2+((60-x)/4)^2`

`A_s= x^2/16+(x^2-120x+3600)/16`

`A_s= (2x^2)/16- 120/16x+3600/16`

`A_s=1/8x^2-15/2x+225`
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This is a quadratic and as the `x^2` term is positive it is of general shape of `uu`

Thus the minimum area (`A_s`) is at the vertex. Let me show you a trick to determining the vertex `x` value.

Write as `A_s=1/8(x^2-(8xx15)/2x)+225`

`A_s=1/8(x^2-color(red)((cancel(8)^4xx15)/(cancel(2)^1))x)+225`

`color(green)("The above is the beginning of the process to 'complete the square'")`

`x_("vertex")= (-1/2)xx(color(red)(-4xx15)) = +30`

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So the 60 cm long wire should be cut so that you have 2 lengths of 30 cm each.