Question

A wire of length 100 centimeters is cut into two pieces. One piece is bent to form a square.

The other piece is bent to form an equilateral triangle. Find the dimensions of the two pieces of wire so that the sum of the areas of the square and the triangle is minimized.

Answer 1

Let $s$ be the side of square and $a$ be the side of equilateral triangle.

Since, the perimeter of square and triangle together formed from the wire of length $100$ cm, therefore,

$4s+3a=100$

$4s+3a-3a=100-3a$

$4s=100-3a$

$s=\frac{100-3a}{4}$

The area of square with side $s$ is $A_S=s^2$

The area of equilateral triangle with side $a$ is $A_T=\frac{\sqrt{3}}{4}a^2$

The sum of the area of both the square and triangle is,

$A=A_S+A_T$

$A=s^2+\frac{\sqrt{3}}{4}a^2$

Substitute $\frac{100-3a}{4}$ for $s$ into above equation to express the area into a single variable $a$ as,

$A(a)=(\frac{100-3a}{4})^2+\frac{\sqrt{3}}{4}a^2$

Differentiate $A(a)$ with respect to $a$ as,

$A'(a)=\frac{d}{da}(\frac{100-3a}{4})^2+\frac{\sqrt{3}}{4}\frac{d}{da}(a^2)$

$=2(\frac{100-3a}{4})\frac{d}{da}(\frac{100-3a}{4})+\frac{\sqrt{3}}{4}(2a)$

$=2(\frac{100-3a}{4})(-\frac{3}{4})+\frac{\sqrt{3}a}{2}$

$=\frac{9a-300}{8}+\frac{\sqrt{3}a}{2}$

$=\frac{9a-300+4\sqrt{3}a}{8}$

$=\frac{a(9+4\sqrt{3})-300}{8}$

Equate $A'(a)$ to zero in order to find the critical value as,

$A'(a)=0$

$\frac{a(9+4\sqrt{3})-300}{8}=0$

$a(9+4\sqrt{3})-300=0$

$a=\frac{300}{9+4\sqrt{3}}$

Plug $\frac{300}{9+4\sqrt{3}}$ for $a$ into relation $s=\frac{100-3a}{4}$ and solve for $s$ as,

$s=\frac{100-3(\frac{300}{9+4\sqrt{3}})}{4}$

$=\frac{900+400\sqrt{3}-900}{4(9+4\sqrt{3})}$

$=\frac{400\sqrt{3}}{4(9+4\sqrt{3})}$

$=\frac{100\sqrt{3}}{9+4\sqrt{3}}$

Here, the second derivative $A"(a)=\frac{9+4\sqrt{3}}{8}>0$ , so total area is minimum.

Therefore, the dimensions of the two pieces are:

Sides of square is: $s=\frac{100\sqrt{3}}{9+4\sqrt{3}}\approx10.874$ cmSides of equilateral triangle is: $a=\frac{300}{9+4\sqrt{3}}\approx18.834$ cm