Posted by Andy G on Wednesday, January 22, 2014 at 8:00pm.
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 = 100-x
side of square = x/4
area of square = x^2 /16
perimeter of circle = 100-x
perimiter of circle = 2πr
2πr = 100-x
r = (100-x)/(2π)
area of circle = π(100-x)^2/(4π^2) = (100-x)^2/(4π)
total area = (1/16)x^2 + (1/(4π))(100-x)^2
domain: clearly 0 < x < 100
Since you labeled your problem as "Pre-Calc" 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 + (25-πx/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 100-44 or 56 cm, same as above.