Q. 22 - Calculus

Question

Find dimensions for each shape in Exercises 21–24 so that the total area enclosed is as large as possible, given that the total edge length is 120 inches. The rounded shapes are half-circles, and the triangles are equilateral.

Step-by-Step Solution

Answer

The dimension for the given shape is $(x,y)=(\frac{120}{\mathrm{\pi }+4},\frac{120}{\mathrm{\pi }+4})$

Step 1. Given Information.

Total edge length$=120 inches$

The rounded shapes are half-circles, and the triangles are equilateral.

Step 2. Find the perimeter of rectangle and circle.

Let x be the diameter of the circle and y be the length of the rectangle.

So,

Perimeter of two circles$=\frac{\mathrm{\pi x}}{2}+\frac{\mathrm{\pi x}}{2}$

                                        $=\mathrm{\pi x}$

Perimeter of the rectangle$=2(x+y)$

                                            $=2x+2y$

Step 3. Find y

So,

$$\begin{gathered} 2x+2y+\mathrm{\pi x}=120 \\ 2\mathrm{y}=120-2\mathrm{x}-\mathrm{\pi x} \\ \mathrm{y}=\frac{120-2\mathrm{x}-\mathrm{\pi x}}{2} \end{gathered}$$

Step 4. Find area of the given figure.

To find the total area enclosed is as large as possible, let the horizontal straight edge have length zero. 

Area, $A=\mathrm{\pi }{\left(\frac{\mathrm{x}}{2}\right)}^2+\mathrm{xy}$

            $$\begin{gathered} =\frac{{\mathrm{\pi x}}^2}{4}+x\left(\frac{120-\mathrm{\pi x}-2\mathrm{x}}{2}\right) \\ =-\frac{{\mathrm{\pi x}}^2}{4}+60x-x^2 \end{gathered}$$

Step 5. Find x.

The area enclosed is as large as possible.

$$\begin{gathered} A=\frac{{\mathrm{\pi x}}^2}{4}+60x-\frac{{\mathrm{\pi x}}^2}{2}-x^2 \\ A\text{'}=\frac{\mathrm{\pi x}}{2}+60-\mathrm{\pi x}-2\mathrm{x} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{\pi x}}{2}+60-\mathrm{\pi x}-2\mathrm{x}=0 \\ -\mathrm{\pi x}-4\mathrm{x}=-120 \\ -\mathrm{x}(\mathrm{\pi }+4)=-120 \\ \mathrm{x}=\frac{120}{\mathrm{\pi }+4} \end{gathered}$$

Step 6. Find y.

Substitute x in the equation, 

$$\begin{gathered} \mathrm{y}=\frac{120-2\mathrm{x}-\mathrm{\pi x}}{2} \\ y=\frac{120-2\left({\displaystyle \frac{120}{\mathrm{\pi }+4}}\right)-\mathrm{\pi }\left({\displaystyle \frac{120}{\mathrm{\pi }+4}}\right)}{2} \\ =\frac{{\displaystyle \frac{120\mathrm{\pi }+480-240-120\mathrm{\pi }}{\mathrm{\pi }+4}}}{2} \\ =\frac{240}{2(\mathrm{\pi }+4)} \\ =\frac{120}{\mathrm{\pi }+4} \end{gathered}$$