Question

A straight piece of wire of 28cm is cut into two pieces. One piece is bent into a square (i.e. dimensions x times x). The other piece is bent into a rectangle with aspect ratio three (i.e. dimensions y times 3y). What are dimensions, in centimeters, of the square and the rectangle such that the sum of their areas is minimized.

Answer 1

Answer on Question #86381 – Math – Geometry

Question

1. A straight piece of wire of $28\mathrm{cm}$ is cut into two pieces. One piece is bent into a square (i.e. dimensions $x$ times $x$). The other piece is bent into a rectangle with aspect ratio three (i.e. dimensions $y$ times $3y$). What are dimensions, in centimeters, of the square and the rectangle such that the sum of their areas is minimized.

Solution

Put $z$-length of the first piece of wire, so length of the second piece is equal to $28 - z$.

Side of the square obtain from the equation

$$4x = z \Rightarrow x = \frac{1}{4} z,$$

area of the square is

$$S_1 = x^2 = \left(\frac{1}{4} z\right)^2 = \frac{z^2}{16}.$$

Sides of the rectangle were obtained from the equation

$$2y + 2 \cdot 3y = 28 - z \Rightarrow y = \frac{1}{8} (28 - z),$$

area of the rectangle is

$$S_2 = y \cdot 3y = \frac{1}{8} (28 - z) \cdot \frac{3}{8} (28 - z) = \frac{3(28 - z)^2}{64}.$$

Overall area is

$$S = S_1 + S_2 = \frac{z^2}{16} + \frac{3(28 - z)^2}{64} = \frac{4z^2 + 3(784 - 56z + z^2)}{64} = \frac{4z^2 + 2352 - 168z + 3z^2}{64} = \frac{7z^2 - 168z + 2352}{64}.$$

So we need find $z$ that minimize $S$. An optimal point can be obtained from the equation $S' = 0$.

$$S' = \left(\frac{7z^2 - 168z + 2352}{64}\right)' = \frac{14z - 168}{64} = 0 \Rightarrow z = 12.$$

$z = 12$ is min cause $S'$ changes its sign from «-» to «+» passing through this point.

In this way

$$\begin{array}{l} x = \frac{12}{4} = 3, \\ y = \frac{1}{8} (28 - 12) = 2. \end{array}$$

Answer: dimension of the square is $3 \times 3 \, \text{cm}$, dimension of the rectangle is $2 \times 6 \, \text{cm}$.

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