Q.Total Revenue is $TR = 60Q - 2Q^2$ and Total Cost is $TC = Q^2 + 8Q + 40$. Find the output level $Q$ that maximises profit, and confirm it is genuinely a maximum.

Profit is Total Revenue minus Total Cost; setting its derivative to zero (equivalently, setting $MR=MC$) locates the profit-maximising output, and the second derivative confirms it is a maximum.

Given: $TR=60Q-2Q^2$, $TC=Q^2+8Q+40$.

Step 1 — Form the profit function: $\pi=TR-TC=(60Q-2Q^2)-(Q^2+8Q+40)=-3Q^2+52Q-40$.

Step 2 — Differentiate and set to zero: $\dfrac{d\pi}{dQ}=-6Q+52$. Setting $-6Q+52=0$ gives $Q=\dfrac{52}{6}=\dfrac{26}{3}\approx8.67$.

Step 3 — Apply the second derivative test: $\dfrac{d^2\pi}{dQ^2}=-6$, which is negative, confirming $Q\approx8.67$ gives a MAXIMUM of profit.

Step 4 — Compute the maximum profit: $\pi(8.67)\approx-3(75.15)+52(8.67)-40\approx-225.45+450.84-40\approx185.4$. …

Stopping after solving $\dfrac{d\pi}{dQ}=0$ without applying the second derivative test — a critical point could be a minimum instead of a maximum; the second derivative test is …