Profit-maximizing price for lawnmowers

Text

The company Barbalbero produces lawnmowers. The production cost is $40$ zinc coins each. At a price of $100$ coins $200$ are sold; at a price of $50$ coins $500$ are sold. Find the price that maximizes Barbalbero’s profit, compute that maximum profit and the number of lawnmowers sold at the optimal price.

Solution

Linear demand curve $q=mp+b$. From the points $(100,200)$ and $(50,500)$: $m=\frac{500-200}{50-100}=-6,\qquad 200=-6\cdot 100+b\Rightarrow b=800\quad\Rightarrow\quad q=800-6p.$ Profit: $\pi(p)=(p-40)(800-6p)=-6p^2+1040p-32000.$ Vertex (maximum): $p^\ast=-\frac{1040}{2\cdot(-6)}=\frac{1040}{12}=\frac{260}{3}\approx 86.67\ \text{coins}.$ Quantity $q=800-6\cdot\dfrac{260}{3}=800-520=280$; profit $\pi_{\max}=\left(\frac{260}{3}-40\right)\cdot 280=\frac{140}{3}\cdot 280=\frac{39200}{3}\approx 13066.67\ \text{coins}.$

$\boxed{p^\ast=\tfrac{260}{3}\approx 86.67,\quad q^\ast=280,\quad \pi_{\max}=\tfrac{39200}{3}\approx 13066.67}$