91Ó°ÊÓ

First, we need to determine the equation for the perimeter of a semicircle and the rectangle and then we need to determine the dimension that will admit the most light.

The figure, $r$represents the radius, $L$represents the length, and $w$represents the width.

The perimeter of the whole window.

$P=w+2L+2\raisebox{1ex}{$\mathrm{Ï€°ù}$}\!\left/ \!\raisebox{-1ex}{$2$}\right..$

$=w+2L+\mathrm{Ï€°ù}.$

$=2r+2L+\mathrm{Ï€°ù}.\left(\mathrm{Since}\mathrm{w}=2\mathrm{r}\right)…\left(1\right).$

The area of the window is given by the equation, role="math" localid="1647339018092" $A=Lw+\frac{{\mathrm{Ï€°ù}}^2}{2}.$

$A=\left(L2r\right)+\frac{{\mathrm{Ï€°ù}}^2}{2}…\left(2\right).$

Put $P=20$in equation (1).

$20=2r+2L+\mathrm{Ï€}r.$

$2L=20-2r-\mathrm{Ï€}r.$

$L=10-r-\frac{\mathrm{Ï€°ù}}{2}…\left(3\right).$

Putting the value of $L$in equation (2).

$A=\left(10-r-\frac{\mathrm{Ï€°ù}}{2}\right)\left(2r\right)+\frac{{\mathrm{Ï€°ù}}^2}{2}.$

$=20r-2r^2-\frac{{\mathrm{Ï€°ù}}^2}{2}.$

$=20r-\left(2+\raisebox{1ex}{$\mathrm{Ï€}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)r^2.$

$=-3.57r^2+20r.$

The $r$coordinate of the vertex of an equation in the form $a{x}^2+bx+c.$

$r=\raisebox{1ex}{$-b$}\!\left/ \!\raisebox{-1ex}{$2a$}\right..$

$=\frac{-20}{2·\left(3.57\right)}.$

$=2.8.$

Substitute $r$inequation (3).

$L=10-r-\frac{\mathrm{Ï€°ù}}{2}.$

$=10-2.8-\frac{\mathrm{Ï€}\left(2.8\right)}{2}.$

$≈2.8ft.$

We know, $w=2r.$

$=2\left(2.8\right).$

$=5.6ft.$