Business Application of Maxima and Minima: formula sheet

Key formulas

Function

y = f(x)

Define the business quantity (profit, cost, revenue) as a function of one decision variable.

First order (necessary) condition

dy/dx = f′(x) = 0

Solve for x to get stationary points. This alone does not tell you maximum or minimum.

Second order condition for a maximum

d²y/dx² < 0 at the stationary point

The curve bends downward. Profit-type functions usually show this.

Second order condition for a minimum

d²y/dx² > 0 at the stationary point

The curve bends upward. Cost-type functions usually show this.

Inconclusive case

d²y/dx² = 0

The test fails. Check the sign of dy/dx on both sides of the point.

Power rule

d(axⁿ)/dx = a·n·xⁿ⁻¹

The constant term differentiates to zero.

Revenue and profit link

Profit = Revenue − Cost; at the optimum, MR = MC

Marginal revenue is dR/dx and marginal cost is dC/dx. This follows from setting d(Profit)/dx = 0.

Total cost

TC = FC + VC(x)

FC is constant. It affects profit but not the profit-maximising output.

Average and marginal cost

AC = TC ÷ x ; MC = d(TC)/dx

Find AC by dividing the whole TC by x before differentiating AC.

Total, average and marginal revenue

TR = p·x ; AR = TR ÷ x = p ; MR = d(TR)/dx

Write p in terms of x from the demand function first.

Linear demand case

p = a − bx ⇒ TR = ax − bx² ; MR = a − 2bx

MR has the same intercept as demand but twice the slope. This holds only for linear demand.

Profit function

π = TR − TC

Use total figures, not per-unit figures, unless the question asks for per-unit profit.

First-order condition for maximum profit

dπ/dx = 0 ⇔ MR = MC

This gives the candidate output. It is not yet proved to be a maximum.

Second-order condition

d²π/dx² < 0 ⇒ maximum ; d²π/dx² > 0 ⇒ minimum

Equivalent to the MC curve cutting MR from below.

Minimum average cost

d(AC)/dx = 0 and d²(AC)/dx² > 0 ⇒ MC = AC

At minimum AC the marginal cost equals average cost.

Maximum revenue

MR = 0 and d²(TR)/dx² < 0

Revenue maximisation gives a different output from profit maximisation unless MC = 0.

MR and price elasticity

MR = p × (1 − 1/|e|)

Here e is the price elasticity of demand. MR is positive only when |e| > 1.

Profit function

π = TR − TC, where TR = P × Q

Write TC and TR in terms of Q before differentiating.

Marginal revenue and marginal cost

MR = d(TR)/dQ; MC = d(TC)/dQ

MC is the derivative of total cost, so fixed cost drops out of MC.

First-order condition for maximum profit

dπ/dQ = 0, which gives MR = MC

This finds the candidate output.

Second-order condition for maximum profit

d²π/dQ² < 0 (equivalently, slope of MR < slope of MC)

Without this check you have not proved a maximum.

Linear demand and its MR

P = a − bQ gives TR = aQ − bQ² and MR = a − 2bQ

MR falls twice as fast as price for a linear demand curve.

Average cost and its minimum

AC = TC ÷ Q; minimum where d(AC)/dQ = 0 and d²(AC)/dQ² > 0

At this point MC = AC.

Per-unit tax or royalty

New TC = old TC + t × Q, so new MC = old MC + t

A lump-sum fixed charge adds to TC but does not change MC, so the profit-maximising output stays the same.

Total inventory cost

TC(Q) = (A ÷ Q) × O + (Q ÷ 2) × C

Add the purchase cost A × P only if price varies with order size. Otherwise it does not affect EOQ.

First-order condition

dTC/dQ = −AO ÷ Q² + C ÷ 2 = 0

Setting this to zero gives the stationary point.

Economic Order Quantity

EOQ = √(2AO ÷ C)

If carrying cost is a percentage i of unit price P, then C = i × P.

Second-order condition

d²TC/dQ² = 2AO ÷ Q³ > 0 for Q > 0

A positive value confirms a minimum.

Minimum total cost

TC at EOQ = √(2AOC)

Excluding purchase cost. Ordering cost equals carrying cost at this point.

Number of orders and cycle

Orders = A ÷ EOQ; Cycle time = EOQ ÷ A (in years)

Multiply cycle time by 365 or 300 as the question states, to get days.

Profit function

π = R − C

R = revenue, C = total cost. Maximise π, not R or C alone, unless the question asks for revenue or cost only.

Condition for maximum

dπ/dq = 0 and d²π/dq² < 0

Equivalent to MR = MC. The second-order check is needed to show it is a maximum.

Per-unit tax on producer

C_new(q) = C(q) + t·q, so MR = MC + t

t is the tax per unit. Use this when the producer bears the tax.

Ad valorem tax on revenue

π = (1 − k)·R − C

k is the tax rate on sales value, written as a decimal.

Lump-sum tax or fixed charge

π_new = π − T

T is a fixed amount. The optimal q does not change.

Linear demand, constant MC, per-unit tax

p = a − b·q; q* = (a − c − t) ÷ (2b)

c is constant marginal cost. The price is p* = a − b·q*.

Revenue maximisation (linear demand)

R = p·q = (a − bq)·q; q = a ÷ (2b)

Revenue peaks where MR = 0. Profit peaks at a lower output whenever MC > 0.

Tax collected

Tax revenue = t × q*

Use the output after the tax, not the output before it.

Equality constraint

Substitute y = g(x) into π(x, y), then solve dπ/dx = 0

Reduces two variables to one.

Quick revision

• Stationary point: first derivative equals zero.
• Second derivative negative: maximum. Positive: minimum.
• Revenue = price × quantity; profit = revenue − total cost.
• Profit is maximised where marginal revenue = marginal cost, provided the second-order condition holds.
• Marginal cost is the derivative of total cost; marginal revenue is the derivative of total revenue.
• Average cost is lowest where it equals marginal cost, given a standard U-shaped curve.
• EOQ = √(2 × annual demand × ordering cost per order ÷ carrying cost per unit per year).
• At EOQ, annual ordering cost equals annual carrying cost in the basic model.
• Number of orders per year = annual demand ÷ EOQ.
• A quantity discount changes the total cost, so compare total cost at each discount level, not just the EOQ.
• Always check that the answer is feasible: non-negative and within any capacity or constraint.
• Finish with a decision statement in rupees or units.

Common mistakes

• Stopping after dy/dx = 0 and calling the point a maximum or minimum. Fix: Always take the second derivative and state its sign. Marks are given for the condition.
• Substituting x into the derivative to find the maximum profit. Fix: The derivative gives the slope, which is zero there. Use the original function for the value.
• Using price p as marginal revenue. Fix: If price falls as x rises, always build TR = p·x first and differentiate it. For p = a − bx, MR = a − 2bx.
• Differentiating AC without dividing TC by x, or confusing AC with MC. Fix: AC = TC ÷ x. MC = dTC/dx. For minimum AC, differentiate AC and set it to zero. Then verify MC = AC.
• Setting price equal to MC instead of MR equal to MC when the firm faces a downward-sloping demand curve. Fix: If demand depends on Q, build TR = P × Q first and differentiate it. Use P = MC only if price is fixed and does not change with Q.
• Skipping the second-order condition. Fix: Always differentiate again, state the sign, and write 'maximum' or 'minimum'. It takes one line and secures marks.
• Using total order cost or carrying cost for the wrong period, such as monthly carrying cost with yearly demand. Fix: Convert A and C to the same period, normally one year, before calculating.
• Taking carrying cost as the average stock times C but using Q instead of Q ÷ 2. Fix: Always write carrying cost as (Q ÷ 2) × C when building the function.
• Adding the tax to revenue or subtracting it from price in the wrong place, which gives a wrong MC. Fix: For a tax paid by the producer, add t·q to total cost. Then MC becomes MC + t.
• Using the pre-tax output to calculate tax collected or profit after tax. Fix: Recompute q from the new profit function. Tax collected = t × new q.

Exam tips

• Show all three parts: first derivative set to zero, second derivative sign, and the final value. Each carries marks in written answers.
• In MCQs, check the sign of the second derivative first. It often eliminates two options at once.
• Read units carefully. Many questions give x in hundreds or ₹ in thousands, and the final answer must be converted.
• Write one line of business meaning, such as produce 3,000 units to earn the highest profit. SCM questions reward a clear recommendation.
• If the problem gives capacity or a minimum order, check the feasible range before you finalise x.
• In MCQs, check the formula first: MR = d(p·x)/dx, not p. Many wrong options are built from using price as MR.
• In written answers, show the second-order test as its own line with the sign. It is an easy mark to protect.
• Put the optimal x back into MR and MC as a check. If they differ, find the error before you write the answer.