Solution — 2019 Question 5 Exponential & Logarithmic Functions| Graphs| Integration & Applications

(a)

When $t = 0,$ $P = 0$ (since the company has just started, profit is zero).

\begin{matrix} k {(1.5)}^{0} - 0.6 & = 0 \\ k - 0.6 & = 0 \\ k & = 0.6 ∎ \end{matrix}

(b)

(c)

When $t = 6,$

\begin{matrix} P = 0.6 \times {1.5}^{6} + (- 0.6) \\ = 6.23 ∎ \end{matrix}

(d)

Using the GC, when $t = 3,$

\begin{matrix} \frac{d P}{d t} & = 0.821 07 \\ = 0.821 ∎ (to 3 s.f.) \end{matrix}

(e)

\begin{matrix} C & = t^{3} - 10 t^{2} + 25 t + 10 \\ \frac{d C}{d t} & = 3 t^{2} - 20 t + 25 \end{matrix}

At stationary points, $\frac{d C}{d t} = 0$

\begin{matrix} 3 t^{2} - 20 t + 25 & = 0 \\ (3 t - 5) (t - 5) & = 0 \end{matrix}

$t = \frac{5}{3} ∎ or t = 5 ∎$

\begin{matrix} \frac{d C}{d t} & = 3 t^{2} - 20 t + 25 \\ \frac{d^{2} C}{d t^{2}} & = 6 t - 20 \end{matrix}

When $t = \frac{5}{3}$

\begin{matrix} \frac{d^{2} C}{d t^{2}} & = 6 (\frac{5}{3}) - 20 \\ = - 10 \end{matrix}

Since $\frac{d^{2} C}{d t^{2}} < 0,$ it is a maximum $∎$ point when $t = \frac{5}{3}$

\begin{matrix} \frac{d^{2} C}{d t^{2}} & = 6 (5) - 20 \\ = 10 \end{matrix}

Since $\frac{d^{2} C}{d t^{2}} > 0,$ it is a minimum $∎$ point when $t = 5$

(f)

(g)

Using GC,

\begin{matrix} \int_{0}^{6} (t^{3} - 10 t^{2} + 25 t + 10) d t & = 114 ∎ \end{matrix}

It represents that the model predicts a total cost of $114$ thousands dollars for this period of six years.