Q. 24 - 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 $(x,y)=(\frac{788-60\sqrt{3}}{13},\frac{30(\sqrt{3}-4)}{13})$

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 triangle and rectangle.

Let $x,y$ be the length and breadth of the triangle respectively, then,

Perimeter of the triangle$=\frac{3y}{3}+\frac{3y}{3}$

                                         $=2y$

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

                                            $=2x+2y$

Step 3. Find the perimeter of whole figure.

$$\begin{gathered} 2y+2x+2y=120 \\ 2x+4y=120 \\ x+2y=60 \\ x=60-2y \end{gathered}$$

Step 4. Find area of the given figure.

$$\begin{gathered} A=2.\frac{\sqrt{3}}{4}{y}^2+xy \\ =\frac{\sqrt{3}}{2}{y}^2+(60-2y)y \\ =\frac{\sqrt{3}}{2}{y}^2+60y-2y^2 \\ A\text{'}=\frac{\sqrt{3}}{2}{y}^2+60y-2y^2 \\ \frac{\sqrt{3}}{2}{y}^2+60y-2y^2=0 \\ \frac{\sqrt{3}}{2}y+60-2y=0 \\ \frac{\sqrt{3}}{2}y-2y=-60 \\ y(\frac{\sqrt{3}}{2}-2)=-60 \\ y=-\frac{60}{{\displaystyle \frac{\sqrt{3}-4}{2}}} \\ =-\frac{30}{\sqrt{3}-4} \\ =-\frac{30}{\sqrt{3}-4}.\frac{\sqrt{3}+4}{\sqrt{3}+4} \\ =-\frac{30(\sqrt{3}-4)}{(-13)} \\ =\frac{30(\sqrt{3}-4)}{13} \end{gathered}$$

Step 5. Find x.

Substitute x in the equation, 

$$\begin{gathered} x=60-2y \\ =60-2\left(\frac{30(\sqrt{3}-4)}{13}\right) \\ =\frac{780-60\sqrt{3}+8}{13} \\ =\frac{788-60\sqrt{3}}{13} \end{gathered}$$