SOLUTION: Minimizing Area) A 24in. piece of string is cut into two pieces. One piece is used to form a circle while the other is used to form a square. How should the string be cut so that t

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Minimizing Area) A 24in. piece of string is cut into two pieces. One piece is used to form a circle while the other is used to form a square. How should the string be cut so that t      Log On


   

Question 410522: Minimizing Area) A 24in. piece of string is cut into two pieces. One piece is used to form a circle while the other is used to form a square. How should the string be cut so that the sum of the areas is a minimum?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
The smallest area is equal to 1013.53155 as shown in the graph below:



the full picture of this graph is shown below:



the solution was derived as follows:

let x = length of string that forms the square.
let 24-x = length of string that forms the circle.

circumference of a square is 4*s where s is equal to 1 side of the square.

since the length of the string forms the circumference of the square, we get:

4*s = x which allows us to derive:

s = (x/4)

area of a square is equal to s^2 which is therefore equal to (x/4)^2 which is therefore equal to x^2/16

area of the square is equal to x^2/16

the circumference of a circle is equal to 2*pi*r

since the length of the remaining string is equal to the circumference of the circle, then we get:

2*pi*r = 24-x

solve for r to get:

r = (24-x) / (2*pi)

the area of a circle is equal to pi * r^2

substitute to get:

area of a circle is equal to pi * ((24-x)^2/(2*pi))^2

simplify this to get:

area of a circle is equal to pi * ((24-x)^2/(4*pi^2)

simplify this further to get:

area of a circle is equal to (24-x)^2 / (4*pi)

simplify this further to get:

area of a circle is equal to (x^2 - 48x + 576) / (4*pi)

let the sum of the area of the square and the circle equal to y.

you get:

y = x^2/16 + ((x^2 - 48x + 576) / (4*pi))

simplify this by finding a common denominator to get:

y = (4*pi*x^2 + 16 * (x^2 - 48*x + 576)) / (4*16*pi)

simplify this further to get:

y = (4*pi*x^2 + 16*x^2 - 768*x + 9216) / (64*pi)

set y = 0 to get:

(4*pi*x^2 + 16*x^2 - 768*x + 9216) / (64*pi) = 0

combine like terms to get:

(4*pi + 16)*x^2 - 768*x + 9216) / 64*pi) = 0

this is your quadratic equation in standard form.

multiply both sides of this equation by (64*pi) to get:

(4*pi + 16)*x^2 - 768*x + 9216) = 0

divide both sides of this equation by 4 to get:

(pi+4)*x^2 - 192*x + 2304 = 0

replace pi with it's constant value of 3.141592654 and combine like terms to get:

7.141592654*x^2 - 192*x + 2304 = 0

in this quadratic equation:

a = 7.141592654
b = -192
c = 2304

max/min point is found by using the formula of x = -b/2a

x = -b/2a becomes:

x = 192/14.28318531 which becomes:

x = 13.44237968

substitute for x in the equation of:

(4*pi + 16)*x^2 - 768*x + 9216) to get:

when x = 13.44237968, y equals 1013.53155

that's the minimum point of your quadratic equation which is modelling the sum of the areas of the square and the circle.