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

## The other piece is bent into a square.

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Mar 29, 2018

$\frac{270}{17}$ cm and $\frac{240}{17}$ cm

#### Explanation:

Let the width of the rectangle be $x$ cm. Then the length is $2 x$ cm, making the perimeter of the rectangle $6 x$ cm. This leaves $\left(30 - 6 x\right)$ cm for the square, whose side will have to be $\frac{3}{2} \left(5 - x\right)$ cm.

The total area now is $A {\text{ cm}}^{2}$ where

$A = 2 {x}^{2} + {\left(\frac{3}{2} \left(5 - x\right)\right)}^{2} = 2 {x}^{2} + \frac{9}{4} {\left(5 - x\right)}^{2}$

so that

$\frac{\mathrm{dA}}{\mathrm{dx}} = 4 x + \frac{9}{4} \times 2 \left(5 - x\right) \times \left(- 1\right) = 4 x - \frac{9}{2} \left(5 - x\right)$

To maximize the area, we must set $\frac{\mathrm{dA}}{\mathrm{dx}} = 0$, leading to

$4 x - \frac{9}{2} \left(5 - x\right) = 0 \implies \left(4 + \frac{9}{2}\right) x = \frac{9}{2} \times 5$
$\implies \frac{17}{2} x = \frac{45}{2} \implies x = \frac{45}{17}$

So, the length of the two pieces must be $\frac{45}{17} \times 6 = \frac{270}{17}$ cm, and $30 - \frac{270}{17} = \frac{240}{17}$ cm

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