Posted by **Andy G** on Wednesday, January 22, 2014 at 8:00pm.

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.

- Pre-Calc -
**Reiny**, Wednesday, January 22, 2014 at 8:59pm
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.

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