Question

(b) 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 the length of the 1^st part of the wire be x cm and the 2^nd part be y cm

$x+y=100$ .....................(1)

therefore one side of squre is (x/4) and that of triangle y/3

Total area enclosed by the two figures (square and triangle)

$AREA=A1+A2=(X/4)^2+(√3)/4 (y/3)^2=(1/16) x^2+((√3)/36) y^2$

from equation (1), $y=100-x$

$A=(1/16) x^2+((√3)/36) 〖100-x〗^2$

Defferentiating

{let have Area as a function of variable x}

$A(x)=[1/16] x^2+[(√3)/36] (100-x)^2$

A(x) is a continuous on the closed interval (0,10) since

$A'(x)=[1/8]x^2+[(√3)/18](100-x)(-1)$

When $A'(x)=0$

$[1/8]x^2+[(√3/)18] (100-x)(-1)=0$

Which is equivalent to

$[1/8]x+[(√3)/18]x=(100√3)/18$

$x=(800√3)/(18-8√3)$ =43.50 $y=(100-43.50)=56.50$

Therefore for minimized

The length of wire required for the square will be x= 43.50 centimeters and the length required for triangle will be y=56.50 centimeters.