17M.1.AHL.TZ2.H_10

pestleMathematicsAIHLPaper 117M· sl-5-6-stationary-points-local-max-and-minsource ↗

A window is made in the shape of a rectangle with a semicircle of radius $r$ metres on top, as shown in the diagram. The perimeter of the window is a constant P metres.

M17/5/MATHL/HP1/ENG/TZ2/10

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

[4]

a.i.

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

[5]

a.ii.

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

[2]

Markscheme / solution

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]

a.i.

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

a.ii.

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

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

$P = r(4 + \pi )$ M1

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

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

[2 marks]

Examiners’ report

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a.i.

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a.ii.

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