Maria's question at Yahoo! Answers regarding optimization

[MarkFL]

Here is the question:

OPTIMIZATION PROBLEM?

A piece of wire 50 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle

(a) How much of the wire should go to the square to maximize the total area enclosed by both figures?

(b) How much of the wire should go to the square to minimize the total area enclosed by both figures?

Here is a link to the question:

OPTIMIZATION PROBLEM? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Maria,

Let's let the length of the wire be $L$, and the let $S$ be the portion of the wire used to form the square, and let the portion of the wire that is the circumference of the circle be $C$

Each side of the square will be $\dfrac{S}{4}$, and so the area of the square is:

$\displaystyle A_S=\left(\frac{S}{4} \right)^2=\frac{S^2}{16}$

The area of the circle is:

$A_C=\pi r^2$

Now, the circumference of the circle is $C$ and so the radius of the circle is:

$\displaystyle r=\frac{C}{2\pi}$

hence the area of the circle in terms of $L$ and $C$ is:

$\displaystyle A_C=\pi\left(\frac{C}{2\pi} \right)^2=\frac{C^2}{4\pi}$

Adding the two areas, we find the total area of the two shapes is:

$\displaystyle A=A_S+A_C=\frac{S^2}{16}+\frac{C^2}{4\pi}$

To find the absolute extrema, we first observe we must have:

$0\le S\le L$ which also means $0\le C\le L$

So, we will find the local extrema within this interval, and also look at the area at the end-points of the interval as possible absolute extrema as well.

There are 3 ways to find the local extremum, since the area is a quadratic in $S$.

For the first two methods, we wish to have an area function in one variable, and so we may use:

$C=L-S$

and we have:

$\displaystyle A(S)=\frac{S^2}{16}+\frac{(L-S)^2}{4\pi}$

Method 1: find the axis of symmetry

We will write the area function in standard quadratic form:

$\displaystyle A(S)=\frac{4+\pi}{16\pi}S^2-\frac{L}{2\pi}S+\frac{L^2}{4\pi}$

Thus, the axis of symmetry is at:

$\displaystyle S=-\frac{-\frac{L}{2\pi}}{2\left(\frac{4+\pi}{16\pi} \right)}=\frac{4L}{4+\pi}$

Since the parabola opens upward, we know the global minimum is at this value of $S$.

Thus, this is the amount of the wire that should go to the square to minimize the total area. Analysis of the end-points reveals:

$\displaystyle A(0)=\frac{L^2}{4\pi}$

$\displaystyle A(L)=\frac{L^2}{16}$

Since $A(0)>A(L)$, we find that to maximize the area, none of the wire should go to the square, and this makes sense as a circle will enclose more area per perimeter than a square.

Method 2: Equate derivative of area function to zero and solve for $S$ to get critical value.

$\displaystyle A(S)=\frac{S^2}{16}+\frac{(L-S)^2}{4\pi}$

$\displaystyle A'(S)=\frac{S}{8}-\frac{L-S}{2\pi}=\frac{\pi S-4(L-S)}{8\pi}=0$

This implies:

$\pi S-4(L-S)=0$

$S=\dfrac{4L}{4+\pi}$

The remaining analysis is the same as for Method 1.

Method 3: Use optimization with constraint

We have the objective function:

$\displaystyle A(C,S)=\frac{S^2}{16}+\frac{C^2}{4\pi}$

subject to the constraint:

$g(C,S)=C+S-L=0$

Using Lagrange multipliers, we find:

$\displaystyle \frac{C}{2\pi}=\lambda$

$\displaystyle \frac{S}{8}=\lambda$

this implies:

$\displaystyle \frac{C}{2\pi}=\frac{S}{8}$

$\displaystyle C=\frac{\pi S}{4}$

Substituting this into the constraint, we find:

$\dfrac{\pi S}{4}+S-L=0$

Solving for $S$, we find:

$S=\dfrac{4L}{4+\pi}$

The remaining analysis is the same as for Method 1.

In conclusion, we have found to minimize the area, we need:

$S=\dfrac{4L}{4+\pi}$

Using the given value $L=50\text{ m}$ we find then:

$S=\dfrac{200}{4+\pi}\,\text{m}$

To maximize the area, we need:

$S=0\text{ m}$.

To Maria and other visitors viewing this topic, I invite you to post other calculus problems here:

Calculus