Q.A manufacturer's profit function is $P(x) = -2x^2 + 40x - 150$ (in thousand rupees), where $x$ is the number of units produced (in hundreds). Find the output that maximises profit, and the maximum profit.

Differentiate the profit function, set it to zero to find the critical output, then confirm a maximum using the second derivative.

$x=10$. $P(10)=-200+400-150=50$.

Step 1 — Differentiate. $P'(x) = -4x+40$.

Step 2 — Set $P'(x)=0$. $-4x+40=0 \Rightarrow x=10$.

Step 3 — Confirm a maximum. $P''(x) = -4$, which is negative for every $x$, so $x=10$ is indeed a maximum.

Step 4 — Find the maximum profit. $P(10) = -2(100)+40(10)-150 = -200+400-150=50$.

Independent check. Evaluate $P(x)$ at values just on either side of $x=10$: $P(9) = -2(81)+360-150=-162+360-150=48$; $P(11)=-2(121)+440-150=-242+440-150=48$. Both neighbouring values ($48$) are indeed lower than $P(10)=50$, directly confirming $x=10$ is a local maximum without relying on the derivative test alone.

Forgetting that $x$ is stated to be measured in hundreds of units and reporting the answer as '10 units' instead of '1,000 units' is an easy units slip in this kind of scaled word problem — always re-read what unit the variable itself represents before stating the final answer.