Question

A piece of wire 25 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

Answer:

10.87 m

Explanation:

Total length of the wire = 25 m

Let the length of one piece is y and other piece is 25 - y

Let the side of square is a.

So, 4 a = y

a = y / 4

And the side of triangle is b

3 b = (25 - y)

b = (25 - y) / 3

Area of square, A1 = side x side =

A1 = y² / 16

Area of equilateral triangle, A2 = $\frac{\sqrt{3}}{4}\times b^{2}$

$A_{2}=\frac{\sqrt{3}}{4}\frac{\left ( 25-y \right )^{2}}{9}$

Total area, A = A1 + A2

$A=\frac{y^{2}}{16}+\frac{\sqrt{3}}{4}\frac{\left ( 25-y \right )^{2}}{9}$

For maxima and minima, fins dA /dy

$\frac{dA}{dy}=\frac{y}{8}-\frac{1.732}{18} \times \frac{25-y}{1}$

It is equal to zero.

$\frac{y}{8}=\frac{1.732}{18} \times \frac{25-y}{1}$

9y = 173.2 -6.928 y

15.928 y = 173.2

y = 10.87 m

So, the length of wire to make square is 10.87 m.