Solution — 2022 Question 5 Graphs| Differentiation & Applications

(a)

\begin{matrix} Total profit \\ = 500 x - (\frac{3}{4} x^{2} + 350 x + 300) \\ = 150 x - \frac{3}{4} x^{2} - 300 ∎ \end{matrix}

\begin{matrix} P & = \frac{(150 x - \frac{3}{4} x^{2} - 300)}{x} \\ = 150 - \frac{3}{4} x - \frac{300}{x} ∎ \end{matrix}

(b)

(c)

\begin{matrix} P & = 150 - \frac{3}{4} x - \frac{300}{x} \\ \frac{d P}{d x} & = - \frac{3}{4} + 300 x^{- 2} \end{matrix}

At turning point, $\frac{d P}{d x} = 0,$

\begin{matrix} \frac{d P}{d x} & = 0 \\ - \frac{3}{4} + 300 x^{- 2} & = 0 \\ \frac{300}{x^{2}} & = \frac{3}{4} \\ 1200 & = 3 x^{2} \\ 3 x^{2} & = 1200 \\ x^{2} & = 400 \\ x & = 20 \end{matrix}

When $x = 20,$

\begin{matrix} P & = 150 - \frac{3}{4} (20) - \frac{300}{20} \\ = 120 \end{matrix}

Hence the turning point on the graph is $(20, 120) ∎$

(d)

Using the GC, $P \geq 100$ when $6.67 \leq x \leq 60$

Since $x$ is an integer, the possible values of $x$ are

$7 \leq x \leq 60 ∎$

(e)

When $x = 220,$

\begin{matrix} P & = 150 - \frac{3}{4} (220) - \frac{300}{220} \\ = - \frac{180}{11} \end{matrix}

Since $- \frac{180}{11} < 0,$ the original model predicted that we will make a loss when producing $220$ machines.
Since Lan finds that he can now make a profit, the original model is no longer appropriate $∎$