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,

$ ParseError: Can't use function '$' in math mode at position 1: $̲ y = mS + c

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

\begin{align*} m\left( 300 \right) + c &= 1200 \\ 300m + c &= 1200 \tag{1} \end{align*}$$ ParseError: {align*} can be used only in display mode.

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

\begin{align*} m\left( 700 \right) + c &= 0 \\ 700m + c &= 0 \tag{2} \end{align*}$$

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

$m = - 3, c = 2100$

$y = - 3 S + 2100$

\begin{align*} P &= \text{Total Revenue} - \text{Costs} \\ &= Sy - 240\,000 \\ &= S\left( -3S + 2100 \right) - 240\,000 \\ &= -3S^2 + 2100S - 240\,000 \\ &= 2100S - 3S^2 - 240\,000\; \QED \end{align*}$$

(e)

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

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

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