Question 263596
A piece of wire 40cm long is to be cut into two pieces.
 One piece will be bent to form a circle; the other will be bent to form a square.
 Find the lengths of the two pieces that cause the sum of the area of the circle and square to be a minimum.
:
Let x = circumference of the circle
then
(40-x) = perimeter of the square
:
Find the radius of the circle
r = {{{x/(2pi)}}}
Find the area of the circle
a = {{{pi(x/(2pi))^2 }}}
a = {{{pi(x^2/(4pi^2))}}}
Cancel pi
a = {{{x^2/(4pi)}}}
:
Find the area of the square:
a = {{{((40-x)/4)^2}}}
a = {{{(x^2 - 80x + 1600)/16}}}
:
:
Total area
A = {{{x^2/(4pi)}}} + {{{(x^2 - 80x + 1600)/16}}}
Change to decimals, easier to combine like terms
A = .07958x^2 + .0625x^2 - 5x + 100
A = .14208x^2 - 5x + 100
:
Find the axis of symmetry [x = -b/(2a)]
x = {{{(-(-5))/(2*.14208)}}}
x = {{{5/.28416}}}
x ~ 17.6 inches, the piece of wire creating a circle
and
40 - 17.6 = 22.4 inches, the piece of wire creating the square
:
These lengths should give minimum area of the circle and square together