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:

Wire used for square is 7.55 m.

Step-by-step explanation:

Let the length of wire cut for square be 'x'.

Given:

Total length of the wire = 25 m

Length of wire for square = 'x'

Length of wire for equilateral triangle = $25-x$

Now, area of square (a) = $x^2$

Area of equilateral triangle (e) is given as:

$e=\frac{\sqrt3}{4}(25-x)^2$

Now, total area (A) is given as:

$A=a+e\\\\A=x^2+\frac{\sqrt3}{4}(25-x)^2$

Now, in order to maximize 'A', the derivative of area with respect to 'x' must be 0.

∴ $\frac{dA}{dx}=0$

$\frac{d}{dx}(x^2+\frac{\sqrt3}{4}(25-x)^2)=0\\\\2x-\frac{\sqrt3}{2}(25-x)=0\\\\2x+\frac{x\sqrt3}{2}=\frac{25\sqrt3}{2}\\\\x(2+\frac{\sqrt3}{2})=\frac{25\sqrt3}{2}\\\\x(4+\sqrt3)=25\sqrt3\\\\x=\frac{25\sqrt3}{4+\sqrt3}=7.55\ m$

Therefore, the length of the wire that is cut for square should be 7.55 m for maximum area.