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:
http://www.mathhelpboards.com/f10/
