Q. 53 - Calculus

Question

Suppose you have a $10$-inch length of wire that you wish to cut and form into shapes. In the given below question, you will determine how to cut the wire to minimize or maximize the area of the resulting shapes.

        Suppose you wish to make one cut in the wire and use the two pieces to form a circle and an equilateral triangle. Determine how to cut the wire so that the combined area of these two shapes is (a) as small as possible and (b) as large as possible. 

Step-by-Step Solution

Answer

Ans:  

(a)   The smallest area occurs when $0.599$ inches are used to make the circle.

(b)   The maximum area is possible when the whole wire is used to make the circle.  

Step 1. Given information.

given,

   The length of the wire is, $10$ inch.

   One cut is made in the wire and a circle and an equilateral are formed.

Step 2. The objective is to determine how to cut the wire so that the combined area is small as possible.

Let, the radius of the circle be $r$ inches, and the sides of an equilateral triangle be  inches

So,

   $\begin{array}{r}P=2\mathrm{\pi r}+3t\\ 10=2\mathrm{\pi r}+3t\\ \end{array}$

Solving the equation for $r$,

    $$\begin{gathered} 10=2\mathrm{\pi r}+3\mathrm{t} \\ \mathrm{t}=\frac{10-2\mathrm{\pi r}}{3} \end{gathered}$$

Step 3. The area is,

      $\begin{array}{r}A=\pi (r{)}^2+\frac{\sqrt{3}}{4}(t{)}^2\\ A=\pi r^2+\frac{\sqrt{3}}{4}{\left(\frac{10-2\pi r}{3}\right)}^2\\ A=\pi r^2+\frac{\sqrt{3}}{4}\frac{1}{9}(10-2\pi r{)}^2\\ A=\pi r^2+0.048\left(100-40\pi r+4{\pi }^2{r}^2\right)\\ A=\pi r^2+4.8-6.032r+1.895r^2\\ A=5.037r^2+4.8-6.032r\end{array}$

Finding its derivatives,  

      $\begin{array}{r}{A}^{\mathrm{\prime }}\left(t\right)=2\left(5.037\right)r-6.032=10.074r-6.032\\ A^{\prime \prime }\left(x\right)=10.074\end{array}$

Equating $10.074r-6.032 to 0$.

     $$\begin{gathered} 10.074r-6.032=0 \\ r=0.599 \end{gathered}$$

Therefore, The smallest area occurs when $0.599$ inches are used to make the circle. $\begin{array}{r}{A}^{\mathrm{\prime }}\left(t\right)=2\left(5.037\right)r-6.032=10.074r-6.032\\ A^{\prime \prime }\left(x\right)=10.074\end{array}$

Step 4. (b) The objective is to determine how to cut the wire so that the combined area is large as possible.

The area of the circle using $10$ inches of wire is,

   $\mathrm{\pi }{\left(\frac{5}{\mathrm{\pi }}\right)}^2=\frac{25}{\mathrm{\pi }}=7.9578$

The area of the triangle with $10$inches of wire is,

     $\frac{\sqrt{3}}{4}{\left(\frac{10}{3}\right)}^2 =4.8112$

The maximum area is possible when the whole wire is used to make the circle.