Solution — 2021 Question 5 Exponential & Logarithmic Functions| Graphs| Simultaneous Linear Equations| Differentiation & Applications

(a)

Let $i,$ $t$ and $p$ denote the amount spent (in dollars) on internet, television and poster advertising respectively.

\begin{matrix} i + t + p & = 220 000 & (1) \end{matrix}

total spending equation

\begin{matrix} t & = 2 i \\ 2 i - t & = 0 & (2) \end{matrix}

television is twice internet

\begin{matrix} 3 p & = t - 11 000 \\ t - 3 p & = 11 000 & (3) \end{matrix}

three times poster is 11,000 less than television

Solving equations (1), (2) and (3) with a GC,

\begin{matrix} i = 61 000, t = 122 000, p = 37 000 \end{matrix}

solution to the system of equations

\begin{matrix} Amount spent on poster advertising & = $ 37 000 ∎ \end{matrix}

required answer for poster advertising

(b)

When $t = 4,$

\begin{matrix} S = 1.5 + 0.8 \ln (2 (4) + 1) \\ = 3.26 ∎ \end{matrix}

This means that in the year up to time $t = 4$ years (i.e. the 4th year), the annual sales were approximately $$ 3.26$ million.

(c)

(d)

\begin{matrix} S & = 1.5 + 0.8 \ln (2 t + 1) \\ \frac{d S}{d t} & = 0.8 (\frac{2}{2 t + 1}) \\ = \frac{1.6}{2 t + 1} ∎ \end{matrix}

(e)

Since $2 t + 1 > 0$ for $t \geq 1,$ we have $\frac{1.6}{2 t + 1} > 0$ for $t \geq 1.$

This means that $S$ is always increasing for $t \geq 1,$ i.e. the model predicts that the annual sales will always be increasing over time. $∎$