Question

A piece of wire 22 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (a) How much wire should be used for the square in order to maximize the total area

Answer 1

9.56 m of the wire should be used for the square, to maximize the total area of the wire

Let x be the side length of the square.

So, the perimeter of the square is:

$\mathbf{P_s=4x}$

Let y represent the side length of the equilateral triangle.

So, the perimeter of the triangle is:

$\mathbf{P_t=3y}$

The length of the wire is given as:

$\mathbf{P=22}$

This implies that:

$\mathbf{P_s + P_t=22}$

Substitute values for Ps and Pt

$\mathbf{4x + 3y=22}$

Make y the subject

$\mathbf{y=\frac{22 -4x}3}$

For an equilateral triangle of side length y, the height (h) of the triangle is:

$\mathbf{h =\frac y2\sqrt 3}$

The area of the triangle is then calculated as:

$\mathbf{A_t = \frac 12 yh}$

This gives

$\mathbf{A_t = \frac 12 y\times \frac y2\sqrt 3}$

$\mathbf{A_t = \frac{y^2}{4}\sqrt 3}$

The area of the square is:

$\mathbf{A_s = x^2}$

So, the total area is:

$\mathbf{A = A_s + A_t}$

$\mathbf{A = x^2 + \frac{y^2}{4}\sqrt 3}$

Substitute $\mathbf{y=\frac{22 -4x}3}$

$\mathbf{A = x^2 + (\frac{22 - 4x}{3})^2 \times \frac{\sqrt 3}{4}}$

Differentiate

$\mathbf{A' = 2x + \frac{\sqrt 3}{4} \times \frac 19 \times 2(22 - 4x) \times (-4)}$

$\mathbf{A' = 2x - \sqrt 3 \times \frac 19 \times 2(22 - 4x) }$

$\mathbf{A' = 2x - \frac{2\sqrt 3}9 (22 - 4x) }$

Set to 0

$\mathbf{2x - \frac{2\sqrt 3}9 (22 - 4x) = 0}$

Rewrite as:

$\mathbf{2x = \frac{2\sqrt 3}9 (22 - 4x) }$

Divide through by 2

$\mathbf{x = \frac{\sqrt 3}9 (22 - 4x) }$

Multiply through by 9

$\mathbf{9x = \sqrt 3 (22 - 4x) }$

Open bracket

$\mathbf{9x = 22\sqrt 3 - 4x\sqrt 3 }$

Collect like terms

$\mathbf{9x +4x\sqrt 3= 22\sqrt 3 }$

Factor out x

$\mathbf{x(9 +4\sqrt 3)= 22\sqrt 3 }$

Solve for x

$\mathbf{x= \frac{22\sqrt 3}{9 +4\sqrt 3} }$

Simplify

$\mathbf{x= \frac{38.11}{9 +6.93} }$

$\mathbf{x= \frac{38.11}{15.93} }$

$\mathbf{x= 2.39 }$

Recall that, the perimeter of the square is:

$\mathbf{P_s=4x}$

So, we have:

$\mathbf{P_s=4 \times 2.39}$

$\mathbf{P_s=9.56}$

Hence, 9.56 m of the wire should be used for the square

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