Maxima, minima and optimisation
At a maximum or minimum the slope is zero, so set dy/dx to zero. A negative second derivative means a maximum; a positive one means a minimum.
Optimisation in five steps
Tests and business results
Find and classify stationary points
Find and classify the stationary points of y = x³ − 6x² + 9x + 2.
- dy/dx = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3).
- dy/dx = 0 gives x = 1 or x = 3.
- d²y/dx² = 6x − 12. At x = 1: −6 < 0, a maximum. At x = 3: +6 > 0, a minimum.
- y(1) = 1 − 6 + 9 + 2 = 6. y(3) = 27 − 54 + 27 + 2 = 2.
Maximum at (1, 6); minimum at (3, 2).
Maximise profit
A factory has R(x) = 150x − 0.2x² and C(x) = 2,000 + 40x + 0.05x². Find the output that maximises profit, the maximum profit and the price.
- P(x) = R − C = 150x − 0.2x² − 2,000 − 40x − 0.05x² = 110x − 0.25x² − 2,000.
- P′(x) = 110 − 0.5x = 0, so x = 220.
- P″(x) = −0.5 < 0, so this is a maximum.
- P(220) = 110(220) − 0.25(220²) − 2,000 = 24,200 − 12,100 − 2,000 = 10,100.
- Check MR = MC: MR = 150 − 0.4(220) = 62 and MC = 40 + 0.1(220) = 62. ✓
- Price = R ÷ x = 150 − 0.2(220) = 106.
Produce 220 units, sell at Rs. 106; maximum profit Rs. 10,100.
Minimise average cost
Total cost is C(x) = 2x² + 50x + 1,800. Find the output with the lowest average cost.
- AC = C ÷ x = 2x + 50 + 1,800/x.
- AC′ = 2 − 1,800/x² = 0, so x² = 900 and x = 30.
- AC″ = 3,600/x³ > 0 at x = 30, so it is a minimum.
- AC(30) = 60 + 50 + 60 = 170.
- Check: MC = 4x + 50 = 170 at x = 30, equal to AC. ✓
30 units, with minimum average cost Rs. 170.
Economic order quantity (EOQ)
A supermarket sells 12,000 bags of rice a year. Each order costs Rs. 2,500 to place, and holding one bag for a year costs Rs. 60. How many bags should it order at a time?
- Total cost TC(Q) = (12,000/Q) × 2,500 + (Q/2) × 60 (ordering + holding).
- TC′(Q) = −30,000,000/Q² + 30 = 0, so Q² = 1,000,000 and Q = 1,000.
- TC″ = 60,000,000/Q³ > 0, so this is a minimum. (Same as EOQ = √(2 × 12,000 × 2,500 ÷ 60) = 1,000.)
- At Q = 1,000: ordering cost 12 × 2,500 = 30,000; holding cost 500 × 60 = 30,000.
Order 1,000 bags at a time (12 orders a year); total cost Rs. 60,000 a year.
A supermarket chain such as Cargills deciding how much rice to order at a time balances ordering cost against storage cost. Setting the derivative of total cost to zero gives the economic order quantity.
Structured question: “Find and classify the stationary points of y = x³ − 3x² − 9x + 5.” y′ = 3(x − 3)(x + 1); max at (−1, 10), min at (3, −22).
Structured question: “Find the EOQ when D = 2,000 units a year, ordering cost Rs. 400 and holding cost Rs. 10 a unit.” √(2 × 2,000 × 400 ÷ 10) = √160,000 = 400 units.
Essay: “R = 150x − 0.2x², C = 2,000 + 40x + 0.05x². Find the profit-maximising output, price and profit, and show MR = MC there.” x = 220, price Rs. 106, profit Rs. 10,100; MR = MC = 62.
Max/min recipe: differentiate, set to zero, solve, test with f″. Negative f″ = frown (maximum); positive f″ = smile (minimum).