Q.A firm's demand function is $p = 200 - 4x$ (price $p$ in rupees when $x$ units are demanded), so the revenue function is $R(x) = p\cdot x = 200x - 4x^{2}$. Find the output level $x$ that maximizes revenue, and verify it is a maximum.

Set $R'(x)=0$ to find the critical output level, then use the second-derivative test to confirm it is a maximum.\n> [!TLDR]\n> $R'(x)=200-8x=0 \Rightarrow x=25$; $R''(x)=-8<0$ confirms a maximum.\n\nMaximum revenue occurs at $x= …

Step 1 — First derivative. $R(x)=200x-4x^2$, so by the power rule term by term:

Step 2 — Critical point. Set $R'(x)=0$:

Step 3 — Classify using the second derivative. Differentiate $R'(x)=200-8x$ once more:

Since $R''(x)=-8<0$ (negative, and constant, so true at $x=25$ as everywhere), by the second-derivative test (§9), $x=25$ is a maximum, not a minimum.

Step 4 — Maximum revenue value. Substitute $x=25$ into $R(x)$:

So revenue is maximized at an output of 25 units, giving a maximum revenue of ₹2,500. …

Nearby-value comparison (shown above): $R(24)=2496$ and $R(26)=2496$, both below $R(25)=2500$, independently confirming $x=25$ is a local maximum …

A common mistake is stopping at $R'(x)=0$ and declaring $x=25$ a maximum WITHOUT checking the second derivative's sign — a critical point could equally be a minimum (if $R''(x)>0$) or neither; the second-derivative …