Smartphones

Optimisation

For maximum revenue, we have to maximise $R=(450-5x)(500+10x)$ . (X= no. Of smartphones sold) $\implies \dfrac {dR_{max}}{dx}=0 \implies x=20$ (Take derivative using product rule or otherwise. ) Thus cost of smartphone = $450-100=350.$

We assume that there is a linear dependency between the price and the number of smartphones sold (n(p)), so n(p) = k p + b. "decrease of $5 on the price will result in the sale of an additional 10 smartphones" gives n(p - 5) - n(p) = 10, i.e. k (p - 5) + b - (k p + b) = -5k = 10, so k = -2 and f(p) = -2 p + b To find b, we use the fact that "shop sells smartphones a week for each", i.e. -2 450 + b = 500, so b = 1400 and f(p) = -2 p + 1400

Revenue is R(p) = n(p) * p = -2 p^2 + 1400p. $\frac{dR}{dp}$ = -4 p + 1400 = 0, so p = 350 is a critical point. Second derivative R'' = -4 < 0 so it is a maximum.

Answer: 350$.