91Ó°ÊÓ

Total edge length$=120inches$

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

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

So,

Perimeter of two circles$=\frac{\mathrm{Ï€³\ae }}{2}+\frac{\mathrm{Ï€³\ae }}{2}$

$=\mathrm{Ï€³\ae }$

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

$=2x+2y$

$$\begin{gathered} 2x+2y+\mathrm{Ï€³\ae }=120 \\ 2\mathrm{y}=120-2\mathrm{x}-\mathrm{Ï€³\ae } \\ \mathrm{y}=\frac{120-2\mathrm{x}-\mathrm{Ï€³\ae }}{2} \end{gathered}$$

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

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

$$\begin{gathered} =\frac{{\mathrm{Ï€³\ae }}^2}{4}+x\left(\frac{120-\mathrm{Ï€³\ae }-2\mathrm{x}}{2}\right) \\ =-\frac{{\mathrm{Ï€³\ae }}^2}{4}+60x-x^2 \end{gathered}$$

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

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

$$\begin{gathered} =\frac{{\mathrm{Ï€³\ae }}^2}{4}+x\left(\frac{120-\mathrm{Ï€³\ae }-2\mathrm{x}}{2}\right) \\ =-\frac{{\mathrm{Ï€³\ae }}^2}{4}+60x-x^2 \end{gathered}$$

The area enclosed is as large as possible.

$$\begin{gathered} A=\frac{{\mathrm{Ï€³\ae }}^2}{4}+60x-\frac{{\mathrm{Ï€³\ae }}^2}{2}-x^2 \\ A\text{'}=\frac{\mathrm{Ï€³\ae }}{2}+60-\mathrm{Ï€³\ae }-2\mathrm{x} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{Ï€³\ae }}{2}+60-\mathrm{Ï€³\ae }-2\mathrm{x}=0 \\ -\mathrm{Ï€³\ae }-4\mathrm{x}=-120 \\ -\mathrm{x}(\mathrm{Ï€}+4)=-120 \\ \mathrm{x}=\frac{120}{\mathrm{Ï€}+4} \end{gathered}$$

Substitute x in the equation,

$$\begin{gathered} \mathrm{y}=\frac{120-2\mathrm{x}-\mathrm{Ï€³\ae }}{2} \\ y=\frac{120-2\left({\displaystyle \frac{120}{\mathrm{Ï€}+4}}\right)-\mathrm{Ï€}\left({\displaystyle \frac{120}{\mathrm{Ï€}+4}}\right)}{2} \\ =\frac{{\displaystyle \frac{120\mathrm{Ï€}+480-240-120\mathrm{Ï€}}{\mathrm{Ï€}+4}}}{2} \\ =\frac{240}{2(\mathrm{Ï€}+4)} \\ =\frac{120}{\mathrm{Ï€}+4} \end{gathered}$$