Q.(a) For the cost function $C = 2x\left(\dfrac{x+5}{x+2}\right) + 7$, prove that Marginal Cost (MC) falls continuously as the output $x$ increases.

(a) $MC=2+\dfrac{12}{(x+2)^{2}}$, whose derivative $-\dfrac{24}{(x+2)^{3}}<0$, so MC falls. (b) 1200 shares, $\unicode{8377}21{,}600$ dividend, $22.5\%$ return.

Part (a) — Marginal cost falls continuously.

Given $C=2x\left(\dfrac{x+5}{x+2}\right)+7=\dfrac{2x^{2}+10x}{x+2}+7$.

Step 1 — Simplify by dividing $2x^{2}+10x$ by $(x+2)$.

$2x^{2}+10x=(2x+6)(x+2)-12\;\Rightarrow\;\dfrac{2x^{2}+10x}{x+2}=2x+6-\dfrac{12}{x+2}.$

So $C=2x+6-\dfrac{12}{x+2}+7=2x+13-\dfrac{12}{x+2}.$

Step 2 — Marginal cost $=\dfrac{dC}{dx}$.

$MC=2-12\cdot\dfrac{d}{dx}(x+2)^{-1}=2-12\cdot\left(-1\right)(x+2)^{-2}=2+\dfrac{12}{(x+2)^{2}}.$

Step 3 — Show MC decreases as $x$ increases.

$\dfrac{d(MC)}{dx}=12\cdot(-2)(x+2)^{-3}=-\dfrac{24}{(x+2)^{3}}<0\quad\text{for }x>0.$

Since the derivative of $MC$ is negative for all $x>0$, $MC$ falls continuously as output $x$ increases (approaching $2$ from above).

Part (b) — Share investment.

Investment $=\unicode{8377}96{,}000$; face value $=\unicode{8377}100$; market price $=\unicode{8377}80$; dividend rate $=18\%$.

(i) Number of shares.

$=\dfrac{\text{Investment}}{\text{Market price}}=\dfrac{96000}{80}=1200\ \text{shares}.$ …