Question

A wire 20 inches long will be cut into two parts, one to be bent to form a square and the other to form a circle. What is the side of the square and the diameter of the circle so that the combined area is least?

Answer 1

Solution:

l = length of wire

cut

a = length for square

b = length for circle

l = a + b = 20in

side of square = $\dfrac{a}{4}$

area = $( \dfrac{a}{4} )^2= \dfrac{a^2}{16}$

perimeter of circle : b = 2$\pi$r

r = $\dfrac{b}{2\pi}$

area = $\pi r^2=\pi ( \dfrac{b}{2\pi})^2= \dfrac{b^2}{4 \pi}$

total area =$\dfrac{a^2}{16}+ \dfrac{b^2}{4\pi}$

a = 20 - b

$\dfrac{(20-b)^2}{16} + \dfrac{b^2}{4\pi}$

$\dfrac{400-40b+b^2}{16}+ \dfrac{16+b^2}{4\pi}$

$25-2.5b+ (\dfrac{1}{16}+ \dfrac{1}{4\pi} )b^2$

derive:

$-\dfrac{10}{4} +( \dfrac{1}{8}+ \dfrac{1}{2\pi} )b$

set to 0 and solve:

$-2.5+( \dfrac{1}{8}+ \dfrac{1}{2\pi} )b=0$

$( \dfrac{1}{8} + \dfrac{1}{2\pi} )b= \dfrac{10}{4}$

$b= \dfrac{( \tfrac{10}{4} )}{( \tfrac{1}{8}+ \tfrac{1}{2\pi} )}$

$b= \dfrac{1}{( \tfrac{1}{20} + \tfrac{1}{5\pi} )}$

b = 8.798

a = 20 - 8.798 = 11.202

a = 11.2

b = 8.8

side of square = $\tfrac{a}{4}=2.8$

diameter of circle = 2r = $2( \tfrac{b}{2\pi})=2.8$