Class 12 Statistics Notes · GSEB
Business Applications
Business Applications — minimise cost and maximise revenue and profit using first and second derivative conditions, with worked Indian business examples. GSEB Class 12 Commerce Statistics notes.
Last updated: 24 Sep 2026
Notes
Optimisation — Where Calculus Meets the Balance Sheet
Everything from the previous topics — power rule, f′ = 0, the f″ test — now works together on one rupee-shaped problem. The function being optimised changes; the method does not.
The one-sentence condition set
The Three Conditions
Minimise Cost C
dC/dx = 0
d²C/dx² > 0
C = 10x² − 1000x + 50000 → x = 50 tons, C″ = 20 > 0 → minimum cost ₹25,000
Maximise Revenue R
dR/dx = 0
d²R/dx² < 0
p = 6000 − 2x → x = 1500 watches, R″ = −4 < 0 → max revenue ₹45,00,000 at p = ₹3,000
Maximise Profit P
dP/dx = 0
d²P/dx² < 0
P = 3x − 100 − 0.015x² → x = 100 units, P″ = −0.03 < 0 → max profit ₹50
MR = MC is the same statement in disguise
Walkthrough: Maximise Watch Revenue
Learn the method first — click through the five steps of a complete Section E/F style problem.
Given — demand and the goal
p = 6000 − 2x (watch demand). Goal: demand x that maximises revenue, and the corresponding price.
R = p · x
Think about it — refrigerator version
Explore: Find the Business Optimum
Now you drive. The four-step finder from Maxima and Minima, wearing business clothes — pick your goal (minimise cost, maximise revenue, or maximise profit), enter the function(s), and step through: differentiate → solve = 0 → f″ test → plug back. The goal point is tagged with a ✓, and any stationary point at negative x is discarded (you cannot produce negative units).
Same four steps as the maxima-minima finder — but the goal depends on the mode: cost target needs f″ > 0.
Revenue and Profit Curves — Build Your Own
Defaults reproduce the watch revenue problem (R = 6000x − 2x²) with a sample cost function. Type your own C(x) and R(x) — the lab builds P = R − C, solves P′ = 0, applies the f″ test, and marks the peaks.
C(x) = 0.015x² + 100 · R(x) = −2x² + 6000x
P(x) = R − C = −2.015x² + 6000x − 100
P′(x) = −4.03x + 6000
Revenue — R = −2x² + 6000x
Profit — P = −2.015x² + 6000x − 100
Revenue check
No feasible stationary point for R(x) in x ≥ 0 — revenue has no interior peak.
Profit check — four-step verdict
P′(x) = −4.03x + 6000 = 0 has no real solution — no stationary profit point.
Read the shapes
Full Worked Problems
Solved Example
Problem
Solution
Minimum cost at x = 50 tons; minimum cost = ₹25,000
Solved Example
Problem
Solution
Maximum revenue at x = 1500 watches, price = ₹3,000 (R = ₹45,00,000)
Solved Example
Problem
Solution
P = 3x − 100 − 0.015x²; maximum profit at x = 100 units (P = ₹50)
Solved Example
Problem
Solution
Maximum profit at x = 200 units; maximum profit = ₹14,000
Exam Condition Cheat-Sheet
| Goal | First condition | Second condition | Equivalently |
|---|---|---|---|
| Minimise cost | dC/dx = 0 | d²C/dx² > 0 | MC = 0 rising |
| Maximise revenue | dR/dx = 0 | d²R/dx² < 0 | MR = 0 (|e| = 1) |
| Maximise profit | dP/dx = 0 | d²P/dx² < 0 | MR = MC |
Negative example — skipping the second derivative
Real businesses run this loop daily
Key Takeaways
Key Takeaways
- ★ Min cost: dC/dx = 0 with d²C/dx² > 0 · Max revenue: dR/dx = 0 with d²R/dx² < 0 · Max profit: dP/dx = 0 with d²P/dx² < 0.
- Profit maximisation is P = R − C — maximise it with the same two-derivative test.
- MR = MC is just dP/dx = 0 written in economics language; both appear in exams.
- Worked template: build the function → differentiate → set to zero → verify with f″ → substitute back for the value.
- So what? Whether it is ₹45 lakh of watch revenue or a chai stall's optimal daily milk order — optimisation is the same four steps with different numbers.