Question

A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the wire should be cut so that the sum of the areas of the square and triangle is minimum?

Answer 1

Let the length of side of square and equilateral triangle be a and r respectively. 

It is given that wire is cut into two parts to form a square and a equilateral triangle 

Therefore, 

perimeter of square + perimeter of equilateral triangle = length of wire

⇒ 4a + 3r = 20

⇒ a = \(\frac{20-3r}{4}\) ... (1)

Let us assume area of square + area of circle = S

(from equation 1)

Condition for maxima and minima,

⇒ \(\frac{dS}{dr}\) = 0

This is the condition for minima 

From equation 1,

a = \(\frac{20-3r}{4}\)

Substituting from equation 2

Hence, side of equilateral triangle and length of square be \(\frac{60}{9+4\sqrt3}\) and \(\frac{20\sqrt3}{9+\sqrt3}\) respectively.