IB Mathematics SL 2.4 Determine key features of graphs AI HL Paper 2- Exam Style Questions- New Syllabus

Question

A cafe makes $x$ litres of coffee each morning. The cafe’s profit each morning, $C$, measured in dollars, is modelled by the following equation:

$C = \frac{x}{10} \left( k^2 – \frac{3}{100} x^2 \right)$
where $k$ is a positive constant.

(a) Find an expression for $\frac{dC}{dx}$ in terms of $k$ and $x$:

(b) Hence find the maximum value of $C$ in terms of $k$. Give your answer in the form $p k^3$, where $p$ is a constant:

(c) (i) The cafe’s manager knows that the cafe makes a profit of $426 when 20 litres of coffee are made in a morning. Find the value of $k$:
(ii) Use the model to find how much coffee the cafe should make each morning to maximize its profit:

(d) Sketch the graph of $C$ against $x$, labelling the maximum point and the x-intercepts with their coordinates:

(e) The manager of the cafe wishes to serve as many customers as possible. Determine the maximum amount of coffee the cafe can make that will not result in a loss of money for the morning:

▶️ Answer/Explanation

Markscheme

(a)
    Rewrite: $C = \frac{x k^2}{10} – \frac{3 x^3}{1000}$
    Differentiate: $\frac{dC}{dx} = \frac{k^2}{10} – \frac{3 \times 3 x^2}{1000}$
    $\frac{dC}{dx} = \frac{k^2}{10} – \frac{9 x^2}{1000}$
Result:
$\frac{k^2}{10} – \frac{9 x^2}{1000}$

(b)
    Set $\frac{dC}{dx} = 0$: $\frac{k^2}{10} – \frac{9 x^2}{1000} = 0$
    $\frac{9 x^2}{1000} = \frac{k^2}{10}$
    $x^2 = \frac{100 k^2}{9}$
    $x = \frac{10 k}{3}$ (since $x > 0$)
    Substitute: $C = \frac{\frac{10 k}{3}}{10} \left( k^2 – \frac{3}{100} \left( \frac{10 k}{3} \right)^2 \right)$
    $\left( \frac{10 k}{3} \right)^2 = \frac{100 k^2}{9}$
    $\frac{3}{100} \times \frac{100 k^2}{9} = \frac{k^2}{3}$
    $C = \frac{k}{3} \left( k^2 – \frac{k^2}{3} \right)$
    $C = \frac{k}{3} \times \frac{2 k^2}{3}$
    $C = \frac{2 k^3}{9}$
Result:
$\frac{2}{9} k^3$

(c)(i)
    Given: $C = 426$ when $x = 20$
    $426 = \frac{20}{10} \left( k^2 – \frac{3}{100} \times 20^2 \right)$
    $426 = 2 \left( k^2 – \frac{3}{100} \times 400 \right)$
    $426 = 2 \left( k^2 – 12 \right)$
    $213 = k^2 – 12$
    $k^2 = 225$
    $k = 15$ (since $k > 0$)
Result:
15

(c)(ii)
    From (b): $x = \frac{10 k}{3}$
    With $k = 15$: $x = \frac{10 \times 15}{3}$
    $x = 50$ litres
Result:
50 litres

(d)
    With $k = 15$: $C = \frac{x}{10} \left( 225 – \frac{3}{100} x^2 \right)$
    X-intercepts: $C = 0$ when $x = 0$ or $225 – \frac{3}{100} x^2 = 0$
    $\frac{3}{100} x^2 = 225$
    $x^2 = 7500$
    $x = \sqrt{7500} \approx 86.6$ litres
    Maximum: At $x = 50$, $C = \frac{50}{10} \left( 225 – \frac{3}{100} \times 2500 \right)$
    $C = 5 \left( 225 – 75 \right) = 750$ dollars
Graph of C against x
Result:
Graph: Cubic shape, (0, 0), (50, 750), (86.6, 0)

(e)
    No loss when $C \geq 0$
    Set $C = 0$: $\frac{x}{10} \left( 225 – \frac{3}{100} x^2 \right) = 0$
    $x = 0$ or $225 – \frac{3}{100} x^2 = 0$
    $\frac{3}{100} x^2 = 225$
    $x^2 = 7500$
    $x = \sqrt{7500} \approx 86.6$ litres
Result:
86.6 litres

IB Mathematics SL 2.4 Determine key features of graphs AI HL Paper 2- Exam Style Questions- New Syllabus

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