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]