Solution — 2018 Question 5 Differentiation & Applications

(a)

Since each order is for $x$ televisions and we will buy $1200$ televisions for the year, we will be making $\frac{1200}{x}$ orders

\begin{matrix} C & = 6 x + 200 (1200) + 50 (\frac{1200}{x}) \\ = 6 x + 240 000 + \frac{60 000}{x} ∎ \end{matrix}

(b)

\begin{matrix} C & = 6 x + 240 000 + \frac{60 000}{x} \\ = 6 x + 240 000 + 60 000 x^{- 1} \\ \frac{d C}{d x} & = 6 - 60 000 x^{- 2} \\ = 6 - \frac{60 000}{x^{2}} \end{matrix}

When $C$ is a minimum, $\frac{d C}{d x} = 0$

\begin{matrix} 6 - \frac{60 000}{x^{2}} & = 0 \\ 6 & = \frac{60 000}{x^{2}} \\ 6 x^{2} & = 60 000 \\ x^{2} & = 10 000 \\ x & = 100 \end{matrix}

\begin{matrix} \frac{d C}{d x} & = 6 - \frac{60 000}{x^{2}} \\ = 6 - 60 000 x^{- 2} \\ \frac{d^{2} C}{d x^{2}} & = 120 000 x^{- 3} \\ = \frac{120 000}{x^{3}} \end{matrix}

When $x = 100,$ $\frac{d^{2} C}{d x^{2}} > 0$ so the value of $C$ is a minimum $∎$

\begin{matrix} Minimum value of C \\ = 6 (100) + 240 000 + \frac{60 000}{100} \\ = 241 200 ∎ \end{matrix}

(c)

This is not a reasonable model as the total cost per year should not be a constant. It should vary depending on the number of televisions bought $∎$

(d)

Let $y$ be the number of sales. Since the graph of the number of sales against the selling price will be a straight line,

$y = m S + c$

When $S = 300,$ $y = 1200,$

\begin{matrix} m (300) + c & = 1200 \\ 300 m + c & = 1200 & (1) \end{matrix}

When $S = 700,$ $y = 0$

\begin{matrix} m (700) + c & = 0 \\ 700 m + c & = 0 & (2) \end{matrix}

Solving (1) and (2) with a GC,

$m = - 3, c = 2100$

$y = - 3 S + 2100$

\begin{matrix} P & = Total Revenue - Costs \\ = S y - 240 000 \\ = S (- 3 S + 2100) - 240 000 \\ = - 3 S^{2} + 2100 S - 240 000 \\ = 2100 S - 3 S^{2} - 240 000 ∎ \end{matrix}

(e)

\begin{matrix} P & = 2100 S - 3 S^{2} - 240 000 \\ \frac{d P}{d S} & = 2100 - 6 S \end{matrix}

At maximum profit, $\frac{d P}{d S} = 0$

\begin{matrix} 2100 - 6 S & = 0 \\ 6 S & = 2100 \\ S & = 350 ∎ \end{matrix}