Question

a piece of wire 11 m long is cut into two pieces. one piece is bent into a square and the other is bent into an equilateral triangle. (round your answers to two decimal places.)
(a) how much wire (in meters) should be used for the square in order to maximize the total area?

m
(b) how much wire (in meters) should be used for the square in order to minimize the total area?

m

Explanation:

Step1: Set up variables

Let \(x\) be the length of wire used for the square. Then the length of wire used for the equilateral triangle is \(11 - x\).
The side length of the square \(s_{1}=\frac{x}{4}\), and its area \(A_{1}=(\frac{x}{4})^{2}=\frac{x^{2}}{16}\).
The side length of the equilateral triangle \(s_{2}=\frac{11 - x}{3}\), and its area \(A_{2}=\frac{\sqrt{3}}{4}(\frac{11 - x}{3})^{2}=\frac{\sqrt{3}}{36}(11 - x)^{2}\).
The total area \(A(x)=\frac{x^{2}}{16}+\frac{\sqrt{3}}{36}(11 - x)^{2}\).

Step2: Find the derivative for part (a) (maximization)

Differentiate \(A(x)\) with respect to \(x\):
\(A^{\prime}(x)=\frac{2x}{16}-\frac{2\sqrt{3}}{36}(11 - x)=\frac{x}{8}-\frac{\sqrt{3}}{18}(11 - x)\).
Set \(A^{\prime}(x) = 0\) for critical points:
\(\frac{x}{8}-\frac{\sqrt{3}}{18}(11 - x)=0\)
\(\frac{9x}{72}+\frac{4\sqrt{3}x}{72}-\frac{44\sqrt{3}}{72}=0\)
\(x(9 + 4\sqrt{3})=44\sqrt{3}\)
\(x=\frac{44\sqrt{3}}{9 + 4\sqrt{3}}\approx11\) (by checking the endpoints: when \(x = 0\), \(A(0)=\frac{\sqrt{3}}{36}\times121\approx5.8\); when \(x = 11\), \(A(11)=\frac{121}{16}\approx7.56\))

Step3: Find the derivative for part (b) (minimization)

We already have \(A^{\prime}(x)=\frac{x}{8}-\frac{\sqrt{3}}{18}(11 - x)\).
Set \(A^{\prime}(x) = 0\):
\(\frac{x}{8}-\frac{\sqrt{3}}{18}(11 - x)=0\)
\(18x-8\sqrt{3}(11 - x)=0\)
\(18x+8\sqrt{3}x=88\sqrt{3}\)
\(x=\frac{88\sqrt{3}}{18 + 8\sqrt{3}}\approx4.80\)

Answer:

(a) \(11.00\) m
(b) \(4.80\) m