Question

A piece of wire 16 in. long is to be cut into two pieces. Let x denote the length of the first piece and 16 − x the length of the second. The first piece is to be bent into a circle and the second piece into a square.

a) Express the total combined area A of the circle and the square as a function of x.

b) For which value of x is the area A a minimum?

c) Using the x-value that you found in part (b), find the ratio of the lengths of the shorter to the longer piece of wire.

Answer 1

Problem number two on the study link given below is similar but not the same as your problem. Study the solution to problem two. Then, post the following about your problem:
1) What is the length of one side of the square in your problem?
2) What is the value of r in the circle of your problem?

Study Link:
http://calclab.math.tamu.edu/~kahlig/151-notes/questUans.pdf

Answer 2

1) the length of one side of the square is (16-x)/4.
2) r=x/2pi

Answer 3

If we bend our first piece into a circle, it will have circumference \(x\); given that we know \(C=2\pi r\), can you determine an expression for its area \(A=\pi r^2\) in terms of \(x\)? A similar question is to be posed for our second piece; the square will have perimeter \(16-x\) -- given we know the perimeter of a square is \(P=4s\), can we determine an expression for its area \(A=s^2\) in terms of \(x\)?

Once you do, realize the sum of these two areas yields your total area in terms of \(x\). Do you remember how to maximize functions? Hint: use Fermat's theorem!
http://en.wikipedia.org/wiki/Fermat%27s_theorem_(stationary_points)

Answer 4

so then, A =\[((16-x)\div4)+(x \div 2 \pi)\]

Answer 5

Neither of those are correct areas.

Answer 6

Sorry, I was at the wrong spot in the problem.
the area of the circle would be:

A = \[\pi (x \div 2\pi )^{2}\]

right? and the area of the square =

\[((16-x)\div 4)^{2}\]

Answer 7

Right, now reduce:$$A_c=\pi\left(\frac{x}{2\pi}\right)^2=\frac{x^2}{4\pi}\\A_s=\left(\frac{16-x}4\right)^2=\frac{256-32x+x^2}{16}=16-2x+\frac{x^2}{16}$$

Answer 8

and then add the two together. Do I have to combine like terms? So that it looks like:
\[(5x ^{2}\div 32)- 2x+16\] ?

Answer 9

Your total area is then \(A=\left(\frac1{4\pi}+\frac1{16}\right)x^2-2x+16\) (after combining like terms), i.e. a quadratic. For a problem like this, we have a parabola; minimizing it involves merely finding the vertex, right? See below for a sketch:
Remembering that the vertex of a parabola is found at \(x=-\frac{b}{2a}\):$$x=-\frac{-2}{2\left(\frac1{4\pi}+\frac1{16}\right)}=\frac1{\frac1{4\pi}+\frac1{16}}=\frac1{\frac{4+\pi}{16\pi}}=\frac{16\pi}{4+\pi}$$