Question
A window is designed in the shape of a semicircle attached to a rectangle.
The rectangular section of the window has dimensions $2x$ metres by $h$ metres.
The window consisting of its two sections is shown in the following diagram.
(a) Let the area of the window be $A$ square metres.
Write down an expression for $A$ in terms of $x$ and $h$.
(b) Let the perimeter of the window be $P$ metres.
Given that $P = 10$, show that $h = \frac{1}{2}(10-2x-\pi x)$.
$$h = \frac{1}{2}(10-2x-\pi x)$$
(c) The window is designed to let in the maximum amount of light.
The rectangular section of the window consists of clear glass and lets in three units of light per square metre.
The semicircular section of the window consists of tinted glass and lets in one unit of light per square metre.
Show that the amount of light, $L$, units, let in by the window is given by $L = 30x – 6x^2 – \frac{5}{2}\pi x^2$.
$$L = 30x – 6x^2 – \frac{5}{2}\pi x^2$$
(d)
(i) Find an expression for $\frac{dL}{dx}$.
(ii) Find the value of $x$ so that the window lets in the maximum amount of light. Justify that this value of $x$ gives a maximum.
(iii) Find the value of $h$ so that the window lets in the maximum amount of light.
▶️Answer/Explanation
Detail Solution
part(a)
Step 1: Identify the area of the rectangular section.
The area of the rectangle is given by:
\( A_{\text{rectangle}} = \text{width} \times \text{height} = 2x \cdot h \).
Step 2: Identify the area of the semicircular section.
The radius of the semicircle is \( x \), so the area of the semicircle is:
\( A_{\text{semicircle}} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi x^2 \).
Step 3: Combine the areas to find the total area \( A \).
Thus, the total area \( A \) is:
\( A = A_{\text{rectangle}} + A_{\text{semicircle}} = 2xh + \frac{1}{2} \pi x^2 \).
The expression for the area of the window is:
Answer: The area \( A \) is \( 2xh + \frac{1}{2} \pi x^2 \).
part(b) Step 1: Identify the perimeter of the window.
The perimeter \( P \) consists of the two vertical sides of the rectangle, the base of the rectangle, and the curved part of the semicircle:
\( P = 2h + 2x + \pi x \).
Step 2: Set the perimeter equal to 10.
Given \( P = 10 \), we have:
\( 2h + 2x + \pi x = 10 \).
Step 3: Rearrange to solve for \( h \).
Rearranging gives:
\( 2h = 10 – 2x – \pi x \).
Dividing by 2 yields:
\( h = \frac{1}{2}(10 – 2x – \pi x) \).
Thus, we have shown that:
Answer: \( h = \frac{1}{2}(10 – 2x – \pi x) \).
part(c) Step 1: Calculate the light from the rectangular section.
The light \( L_{\text{rectangle}} \) from the rectangular section is:
\( L_{\text{rectangle}} = 3 \cdot A_{\text{rectangle}} = 3(2xh) = 6xh \).
Step 2: Calculate the light from the semicircular section.
The light \( L_{\text{semicircle}} \) from the semicircular section is:
\( L_{\text{semicircle}} = 1 \cdot A_{\text{semicircle}} = \frac{1}{2} \pi x^2 \).
Step 3: Combine the light contributions to find total light \( L \).
Thus, the total light \( L \) is:
\( L = L_{\text{rectangle}} + L_{\text{semicircle}} = 6xh + \frac{1}{2} \pi x^2 \).
Step 4: Substitute \( h \) from part (b) into the light equation.
Substituting \( h = \frac{1}{2}(10 – 2x – \pi x) \) into \( L \):
\( L = 6x \left(\frac{1}{2}(10 – 2x – \pi x)\right) + \frac{1}{2} \pi x^2 \).
This simplifies to:
\( L = 3x(10 – 2x – \pi x) + \frac{1}{2} \pi x^2 = 30x – 6x^2 – 3\pi x^2 + \frac{1}{2} \pi x^2 \).
Combining the terms gives:
\( L = 30x – 6x^2 – \frac{5}{2} \pi x^2 \).
Thus, we have shown that:
Answer: \( L = 30x – 6x^2 – \frac{5}{2} \pi x^2 \).
part(d) (i) Step 1: Differentiate \( L \) with respect to \( x \).
Using the expression \( L = 30x – 6x^2 – \frac{5}{2} \pi x^2 \), we differentiate:
\(\frac{dL}{dx} = 30 – 12x – 5\pi x\).
Thus, the expression for \(\frac{dL}{dx}\) is:
Answer: \(\frac{dL}{dx} = 30 – 12x – 5\pi x\).
(d) (ii) Step 1: Set the derivative equal to zero to find critical points.
Setting \( \frac{dL}{dx} = 0 \):
\( 30 – 12x – 5\pi x = 0 \).
Step 2: Solve for \( x \).
Rearranging gives:
\( 12x + 5\pi x = 30 \)
\( x(12 + 5\pi) = 30 \)
\( x = \frac{30}{12 + 5\pi} \).
Step 3: Justify that this value gives a maximum.
To confirm it is a maximum, we check the second derivative:
\(\frac{d^2L}{dx^2} = -12 – 5\pi\).
Since this is negative, it indicates a maximum.
Thus, the value of \( x \) for maximum light is:
Answer: \( x = \frac{30}{12 + 5\pi} \).
(d) (iii) Step 1: Substitute \( x \) back into the expression for \( h \).
Using \( h = \frac{1}{2}(10 – 2x – \pi x) \):
Substituting \( x = \frac{30}{12 + 5\pi} \):
\( h = \frac{1}{2}\left(10 – 2\left(\frac{30}{12 + 5\pi}\right) – \pi\left(\frac{30}{12 + 5\pi}\right)\right) \).
Step 2: Simplify \( h \).
Calculating gives:
\( h = \frac{1}{2}\left(10 – \frac{60}{12 + 5\pi} – \frac{30\pi}{12 + 5\pi}\right) \)
$h=\frac{1}{2}[\frac{120+50\pi -60-30\pi}{12+5\pi}]$
$h=\frac{1}{2}[\frac{60+20\pi}{12+5\pi}]$
$h=\frac{30+10\pi}{12+5\pi}$
————Markscheme—————–
(a) let $A_R$ denote the area of the rectangle and $A_S$ denote the area of the semicircle
one correct area $A_R = 2xh$ OR $A_S = \frac{1}{2}\pi x^2$
$A = 2xh + \frac{1}{2}\pi x^2 = x\left(2h + \frac{1}{2}\pi x\right)$
(b) attempts to find a correct expression for the total perimeter
$(P =) 2x + 2h + \pi x$, base + 2 x height + half circumference
$2x + 2h + \pi x = 10$ OR $2h = 10 – 2x – \pi x$
$h = \frac{1}{2}(10 – 2x – \pi x)$
(c) $L = 3(2xh) + \frac{1}{2}\pi x^2$
substitutes $h = \frac{1}{2}(10 – 2x – \pi x)$ into their expression for $L$
$L = 6x\left(\frac{1}{2}(10 – 2x – \pi x)\right) + \frac{1}{2}\pi x^2$
$= 30x – 6x^2 – 3\pi x^2 + \frac{1}{2}\pi x^2 = \left(30x – \left(6 + \frac{5\pi}{2}\right)x^2\right)$
$L = 30x – 6x^2 – \frac{5}{2}\pi x^2$
(d) (i) $\frac{dL}{dx} = 30 – 12x – 5\pi x = 30 – (12 + 5\pi)x \quad (\text{accept } \frac{dL}{dx} = 30 – 27.7x)$
(ii) $\text{recognition that }$ $\frac{dL}{dx} = 0 \text{ (may be represented graphically)}$
$x = 1.08272…$
$x = 1.08 \left(= \frac{30}{12 + 5\pi}\right) \text{ m}$
$\text{correct reasoning to justify a maximum}$
$L \text{ is a quadratic (function of } x \text{) with a negative coefficient of } x^2$ $\text{ (may be represented as a sketch indicating maximum point) OR}$
$\text{a clearly labelled sign diagram showing the change in gradient OR}$
$\frac{d^2L}{dx^2} = -12 – 5\pi (= -27.7079…) < 0$
(iii) attempts to substitute their value of $x$ into $h$
$h = 2.21654…$
$h = 2.22 \left(=\frac{30+10\pi}{12+5\pi}\right) \text{(m)}$