Find the area of the window in terms of P and $r$.

Find the width of the window in terms of P when the area is a maximum, justifying that this is a maximum.

Show that in this case the height of the rectangle is equal to the radius of the semicircle.

the width of the rectangle is $2r$ and let the height of the rectangle be $h$

$P = 2r + 2h + \pi r$     (A1)

$A = 2rh + \frac{{\pi {r^2}}}{2}$     (A1)

$h = \frac{{{\text{P}} - 2r - \pi r}}{2}$

$A = 2r\left( {\frac{{{\text{P}} - 2r - \pi r}}{2}} \right) + \frac{{\pi {r^2}}}{2}\,\,\,\left( { = \operatorname{P} r - 2{r^2} - \frac{{\pi {r^2}}}{2}} \right)$     M1A1

[4 marks]

$\frac{{{\text{d}}A}}{{{\text{d}}r}} = {\text{P}} - 4r - \pi r$     A1

$\frac{{{\text{d}}A}}{{{\text{d}}r}} = 0$     M1

$\Rightarrow r = \frac{{\text{P}}}{{4 + \pi }}$     (A1)

hence the width is $\frac{{2{\text{P}}}}{{4 + \pi }}$     A1

$\frac{{{{\text{d}}^2}A}}{{{\text{d}}{r^2}}} = - 4 - \pi < 0$     R1

hence maximum     AG

[5 marks]

EITHER

$h = \frac{{{\text{P}} - 2r - \pi r}}{2}$

$h = \frac{{{\text{P}} - \frac{{2{\text{P}}}}{{4 + \pi }} - \frac{{{\text{P}}\pi }}{{4 + \pi }}}}{2}$     M1

$h = \frac{{4{\text{P}} + \pi {\text{P}} - 2{\text{P}} - \pi {\text{P}}}}{{2(4 + \pi )}}$    A1

$h = \frac{{\text{P}}}{{(4 + \pi )}} = r$     AG

OR

$h = \frac{{{\text{P}} - 2r - \pi r}}{2}$

$P = r(4 + \pi )$     M1

$h = \frac{{r(4 + \pi ) - 2r - \pi r}}{2}$     A1

$h = \frac{{4r + \pi r - 2r - \pi r}}{2} = r$     AG

[2 marks]