PreCalc
posted by Andy G on .
A wire 100 centimeters in length is cut into two separate pieces (not necessarily equal). ON epiece is bent to form a square and the other to form a circle. Let x equal the length of the wire used to form the square.
a. Write the function that represents the are of the two figures combined.
b. Determine the domain of the function.
c. Find the value(s) of x that yield a max and minimum area.
d. explain your reasoning.

let me say right away, that defining x to be the length of the wire used for the square is a poor choice. But ...
let the piece used for the square be x
then the piece used for the circle = 100x
side of square = x/4
area of square = x^2 /16
perimeter of circle = 100x
perimiter of circle = 2πr
2πr = 100x
r = (100x)/(2π)
area of circle = π(100x)^2/(4π^2) = (100x)^2/(4π)
total area = (1/16)x^2 + (1/(4π))(100x)^2
domain: clearly 0 < x < 100
Since you labeled your problem as "PreCalc" I can't take the derivative and set it equal to zero, like I want to.
So, expanding and simplifying the area equation I got
Area = (1/16 + 1/(4π))x^2  (50/π)x + 2500/π
which is a parabola opening upwards, the vertex would represent a minimum area
the x of the vertex is b/(2a)
= (50/π) / (2(1/16 + 1/(4π))) = appr 56 cm
so a min total will be obtained when x = 56 cm
clearly the maximum will be obtained if x = 100, that is all is used for a circle.
Here is how I would do this:
let the radius be x, and the side of the square be y
4y + 2πx = 100
4y = 100  2πr
y = 25  πx/2
area = πx^2 + (25πx/2)^2
=πx^2 + 625  25πx + π^2 x^2/4
= (π + π^2/4)x^2  25πx + 625
the x coordinate of this parabola is 25π/(2π + π^2/2)
= 25/(2 + π/2) = appr 7.00
so the length for the circle = 2πx = 44
and the circle uses 10044 or 56 cm, same as above.