Q. 52 - 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 square and an equilateral triangle. Determine how to cut the wire so that the combined area is (a) as small as possible and (b) as large as possible. 

Step-by-Step Solution

Answer

Ans:  

(a)   The smallest area occurs when $5.625$ inches are used to make the triangle.

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

Step 1. Given information.

given,

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

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

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

Let, the side of the square be $s$ inches, and the sides of an equilateral triangle be $t$ inches

So,

    $\begin{array}{r}P=4s+3t\\ 10=4s+3t\\ \end{array}$

Solving the equation for $s$,

    $s=\frac{10-3t}{4}$

Step 3. The area is,

  $\begin{array}{r}A=(s{)}^2+\frac{\sqrt{3}}{4}(t{)}^2\\ A={\left(\frac{10-3t}{4}\right)}^2+\frac{\sqrt{3}}{4}(t{)}^2\\ A=\frac{100}{16}-\frac{60t}{16}+\frac{9t^2}{16}+\frac{\sqrt{3}}{4}(t{)}^2\end{array}$

Finding its derivatives, 

     $\begin{array}{r}{A}^{\mathrm{\prime }}\left(t\right)=-\frac{60}{16}+\frac{18t}{16}+\frac{\sqrt{3}}{4}\left(2\right)\left(t\right)=-\frac{15}{4}+2t\\ A^{\prime \prime }\left(x\right)=\frac{9}{8}+\frac{\sqrt{3}}{2}\approx 2\end{array}$

Equating  $-\frac{15}{4}+2t$ to $0$,

     $$\begin{gathered} -\frac{15}{4}+2t=0 \\ t=\frac{15}{8}=1.875 \end{gathered}$$

The minimum inches required for a triangle is,

       $3(1.875)=5.625$ inches.

Therefore, the smallest area occurs when $5.625$ inches is used to make the triangle.

Step 2. (b) The objective is to determine how to cut the wire so that the combined area is large as possible.

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

        ${\left(\frac{10}{4}\right)}^2=6.25$

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 square.