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Class 11 · Tamil Nadu State Board

Applications of Diff Calculus Solutions | KalviGavani

Samacheer Kalvi 11th Business Maths Applications of Differential Calculus exercise solutions — maxima minima, marginal cost revenue, elasticity of demand, related rates, worked examples, MCQs, full answers.

Updated 15 June 202619 min read

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This page contains complete exercise solutions for Chapter: Applications of Differential Calculus from the Samacheer Kalvi 11th Business Maths textbook (Tamil Nadu State Board).

  • MCQs and short answers — pick the right option or write a one-liner; each solution explains why.
  • Fill-in-the-blanks — the exact word or value to fill, with a brief reason.
  • True or False — the verdict plus a one-sentence correction when false.
  • Match the Following — paired with a line of justification.
  • Detail questions — full step-by-step working inside code blocks so you can copy and practise.

Read the Key Concepts table first for a quick recap, then jump to the section you need.

Key Concepts (Quick Recap)

ConceptFormula / RuleInterpretation
Increasing functionf'(x) > 0 on an intervalFunction rises as x increases
Decreasing functionf'(x) < 0 on an intervalFunction falls as x increases
Critical pointf'(x) = 0 or f'(x) does not existCandidate for max or min
First derivative testSign change of f'(x) around cf' changes + to − → max; − to + → min
Second derivative testf''(c) > 0 → local min; f''(c) < 0 → local maxIf f''(c) = 0, test is inconclusive
Marginal Cost (MC)MC = dC/dxCost of producing one more unit
Marginal Revenue (MR)MR = dR/dxRevenue from selling one more unit
Marginal Profit (MP)MP = dP/dx = MR − MCProfit from one more unit
Average Cost (AC)AC = C(x)/xCost per unit
Average Revenue (AR)AR = R(x)/xRevenue per unit (equals price)
MC = AC at min ACd(AC)/dx = 0 ⟹ MC = ACMC curve crosses AC at its lowest point
MR = MC at max profitdP/dx = 0 ⟹ MR = MCProfit is maximised when MR = MC
Elasticity of Demand (Ed)Ed = −(dQ/dP) × (P/Q)Measures responsiveness of demand to price
Elastic demandEd > 1% change in Q > % change in P
Inelastic demandEd < 1% change in Q < % change in P
Unitary demandEd = 1% change in Q = % change in P
Related ratesdy/dt = (dy/dx) × (dx/dt)Chain rule links rates of change

I. Choose the Best Answer (MCQs)

1. If f'(x) > 0 for all x in (a, b), then f(x) is ______ in (a, b).

  • (A) Constant
  • (B) Decreasing
  • (C) Increasing
  • (D) Neither increasing nor decreasing
Answer(C) Increasing
ExplanationWhen the first derivative is positive over an interval, the function is strictly increasing on that interval.

2. A critical point x = c of f(x) is a local maximum if:

  • (A) f''(c) > 0
  • (B) f''(c) < 0
  • (C) f'(c) > 0
  • (D) f''(c) = 0
Answer(B) f''(c) < 0
ExplanationBy the second derivative test, if f''(c) < 0, the function is concave down at c, so c is a point of local maximum.

3. The marginal cost function is the derivative of:

  • (A) Average cost
  • (B) Total revenue
  • (C) Total cost
  • (D) Profit
Answer(C) Total cost
ExplanationMarginal cost MC = dC/dx, where C(x) is the total cost function.

4. At the point where average cost is minimum:

  • (A) MC = 0
  • (B) MC = AC
  • (C) MC > AC
  • (D) MC < AC
Answer(B) MC = AC
ExplanationDifferentiating AC = C(x)/x and setting it to zero yields the condition MC = AC. This is a key result in business mathematics.

5. If the demand function is Q = 100 − 2P, the elasticity of demand at P = 10 is:

  • (A) 1
  • (B) 1/4
  • (C) 4
  • (D) 2
Answer(B) 1/4

Explanation:

Q = 100 − 2P dQ/dP = −2 At P = 10: Q = 100 − 20 = 80 Ed = −(dQ/dP) × (P/Q) = −(−2) × (10/80) = 2 × 0.125 = 0.25 = 1/4 Since Ed < 1, demand is inelastic at P = 10.

6. Profit is maximum when:

  • (A) MR = 0
  • (B) MC = 0
  • (C) MR = MC
  • (D) MR < MC
Answer(C) MR = MC
ExplanationProfit P(x) = R(x) − C(x). Setting dP/dx = 0 gives dR/dx − dC/dx = 0, i.e., MR = MC. The second-order condition MR < MC confirms it is a maximum.

7. If y = x^2 and x is increasing at the rate of 3 units/s, then the rate at which y is increasing when x = 2 is:

  • (A) 6 units/s
  • (B) 12 units/s
  • (C) 4 units/s
  • (D) 24 units/s
Answer(B) 12 units/s

Explanation:

y = x^2 dy/dt = 2x (dx/dt) At x = 2, dx/dt = 3: dy/dt = 2 × 2 × 3 = 12 units/s

II. Fill in the Blanks

1. If f'(x) ______ 0 for all x in (a, b), then f(x) is a decreasing function on (a, b).

Answer< (less than)

A function is decreasing when its first derivative is negative.

2. The points where f'(x) = 0 are called ______ points.

AnswerCritical (or stationary)

Critical points are values of x where the derivative equals zero or does not exist; they are candidates for local extrema.

3. Marginal revenue is the derivative of ______ with respect to x.

AnswerTotal revenue (R(x))

MR = dR/dx represents the additional revenue earned from selling one more unit.

4. If f''(c) ______ 0, then x = c is a point of local minimum.

Answer> (greater than)

The second derivative test states that f''(c) > 0 implies a local minimum.

5. At the minimum of average cost, marginal cost equals ______.

AnswerAverage cost (AC)

This is a fundamental result: d(AC)/dx = 0 leads to MC = AC at the minimum average cost point.

6. Elasticity of demand Ed = −(dQ/dP) × (P/Q). If Ed = 1, demand is said to be ______.

AnswerUnitary elastic

When Ed = 1, the percentage change in quantity demanded equals the percentage change in price.

7. If Ed > 1, the demand is ______.

AnswerElastic (or relatively elastic)

When the absolute value of elasticity exceeds 1, quantity demanded is highly responsive to price changes.

8. The first derivative test checks the ______ of f'(x) on either side of a critical point.

AnswerSign (or sign change)

The first derivative test examines how the sign of f'(x) changes as we pass through a critical point.

9. Profit P(x) = R(x) − C(x). Profit is maximised when dP/dx = 0, i.e., when ______.

AnswerMR = MC

Setting dP/dx = dR/dx − dC/dx = 0 gives MR = MC, the standard profit-maximisation condition in economics.

10. Average cost AC = C(x)/x. Differentiating AC with respect to x and simplifying yields (______ − AC)/x.

AnswerMC
AC = C/x d(AC)/dx = (x·MC − C)/x^2 = (MC − AC)/x

Setting this to zero gives MC = AC.

11. If f'(x) changes from negative to positive at x = c, then f(c) is a local ______.

AnswerMinimum

When the derivative switches from negative (function decreasing) to positive (function increasing), the function has a local minimum at that point.

12. In related rates problems, we use the ______ rule to connect dy/dt and dx/dt.

AnswerChain

The chain rule gives dy/dt = (dy/dx) × (dx/dt), linking the rate of change of y to the rate of change of x through their functional relationship.

III. True or False

1. If f'(x) > 0 for all x, then f(x) is a decreasing function.

AnswerFalse

Correction: If f'(x) > 0, the function is increasing. A decreasing function requires f'(x) < 0.

2. At the point where profit is maximum, marginal revenue equals marginal cost.

AnswerTrue

This is the standard profit-maximisation condition: dP/dx = 0 implies MR = MC.

3. If f''(c) = 0, then the second derivative test confirms that x = c is a point of inflection.

AnswerFalse

Correction: f''(c) = 0 means the second derivative test is inconclusive. The point may be a maximum, minimum, or inflection — you must use other methods (first derivative test or higher-order derivatives) to decide.

4. When elasticity of demand Ed < 1, a price increase will decrease total revenue.

AnswerFalse

Correction: When Ed < 1 (inelastic demand), a price increase increases total revenue because the percentage drop in quantity is smaller than the percentage rise in price.

5. Marginal cost is the cost of producing one additional unit of output.

AnswerTrue

Marginal cost MC = dC/dx approximates the cost of producing one more unit when production is at level x.

IV. Match the Following

Column AColumn B
1. f'(x) = 0(a) Demand is unitary elastic
2. f''(c) > 0(b) Marginal cost
3. dC/dx(c) Critical point
4. Ed = 1(d) Local minimum
5. MR = MC(e) Profit is maximum
6. Ed > 1(f) Demand is elastic
Answer
  • 1 → (c): f'(x) = 0 identifies a critical point (stationary point).
  • 2 → (d): f''(c) > 0 means the function is concave up, giving a local minimum.
  • 3 → (b): dC/dx is the definition of marginal cost.
  • 4 → (a): Ed = 1 means demand is unitary elastic.
  • 5 → (e): MR = MC is the profit-maximisation condition.
  • 6 → (f): Ed > 1 means demand is elastic (quantity responds more than proportionally to price).

V. Answer Briefly

1. Define marginal cost and marginal revenue. Write their formulas.

Marginal cost is the rate of change of total cost with respect to the number of units produced. Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

Marginal Cost: MC = dC/dx Marginal Revenue: MR = dR/dx

2. State the second derivative test for maxima and minima.

Let f be a twice-differentiable function and let c be a critical point (f'(c) = 0).

If f''(c) > 0 → f(c) is a local minimum If f''(c) < 0 → f(c) is a local maximum If f''(c) = 0 → Test is inconclusive

3. What is elasticity of demand? Give its formula and state the three types.

Elasticity of demand measures the responsiveness of quantity demanded to a change in price.

Ed = −(dQ/dP) × (P/Q) Ed > 1 → Elastic demand (quantity is highly responsive to price) Ed < 1 → Inelastic demand (quantity is less responsive to price) Ed = 1 → Unitary elastic demand (proportional response)

4. State the relationship between marginal cost and average cost at the minimum of average cost.

At the point where average cost is minimum, marginal cost equals average cost.

AC = C(x)/x d(AC)/dx = (MC − AC)/x = 0 Therefore, MC = AC at minimum AC.

This means the marginal cost curve intersects the average cost curve at its lowest point.

VI. Answer in Detail

1. Finding Maxima and Minima — Step-by-Step with Worked Business Example

Procedure to find local maxima and minima:

Step 1: Find f'(x) and set f'(x) = 0 to get critical points. Step 2: Find f''(x). Step 3: Evaluate f''(x) at each critical point. f''(c) > 0 → local minimum at x = c f''(c) < 0 → local maximum at x = c f''(c) = 0 → inconclusive; use first derivative test.

Worked Example — Profit Maximisation:

A firm's profit function is P(x) = −x^3 + 12x^2 + 60x − 500, where x is the number of units produced. Find the output level that maximises profit.

Given: P(x) = −x^3 + 12x^2 + 60x − 500 Step 1 — Find P'(x): P'(x) = −3x^2 + 24x + 60 Set P'(x) = 0: −3x^2 + 24x + 60 = 0 Divide by −3: x^2 − 8x − 20 = 0 (x − 10)(x + 2) = 0 x = 10 or x = −2 Since x ≥ 0 (output cannot be negative), the only critical point is x = 10. Step 2 — Find P''(x): P''(x) = −6x + 24 Step 3 — Evaluate P''(x) at x = 10: P''(10) = −60 + 24 = −36 Since P''(10) = −36 < 0, the function is concave down. Therefore, x = 10 is a point of local maximum. Step 4 — Maximum profit: P(10) = −(10)^3 + 12(10)^2 + 60(10) − 500 = −1000 + 1200 + 600 − 500 = 300 Maximum profit = ₹300 at x = 10 units.

Verification with first derivative test:

For x < 10 (say x = 5): P'(5) = −3(25) + 24(5) + 60 = −75 + 120 + 60 = 105 > 0 For x > 10 (say x = 15): P'(15) = −3(225) + 24(15) + 60 = −675 + 360 + 60 = −255 < 0 P'(x) changes from positive to negative at x = 10. This confirms x = 10 is a local maximum.

2. Calculating Marginal Cost, Marginal Revenue, and Profit Maximisation

A company has the following cost and revenue functions:

Cost function: C(x) = x^3 − 6x^2 + 15x + 240 Revenue function: R(x) = 30x

Find: (i) Marginal cost, (ii) Marginal revenue, (iii) Profit function, (iv) Output for maximum profit, (v) Maximum profit.

(i) Marginal Cost: MC = dC/dx = 3x^2 − 12x + 15 (ii) Marginal Revenue: MR = dR/dx = 30 (iii) Profit Function: P(x) = R(x) − C(x) = 30x − (x^3 − 6x^2 + 15x + 240) = −x^3 + 6x^2 + 15x − 240 (iv) Output for Maximum Profit: P'(x) = −3x^2 + 12x + 15 Set P'(x) = 0: −3x^2 + 12x + 15 = 0 Divide by −3: x^2 − 4x − 5 = 0 (x − 5)(x + 1) = 0 x = 5 or x = −1 Since x ≥ 0, the critical point is x = 5. Second derivative: P''(x) = −6x + 12 P''(5) = −30 + 12 = −18 < 0 Since P''(5) < 0, profit is maximum at x = 5. Verification: At x = 5, MR = 30 and MC = 3(25) − 12(5) + 15 = 75 − 60 + 15 = 30. MR = MC = 30 confirms maximum profit. (v) Maximum Profit: P(5) = −(5)^3 + 6(5)^2 + 15(5) − 240 = −125 + 150 + 75 − 240 = −140 Maximum profit = −₹140 (which is a loss of ₹140). The firm minimises its loss at x = 5; any other output level would produce an even larger loss.

Average cost at x = 5:

AC = C(x)/x = (x^3 − 6x^2 + 15x + 240) / x = x^2 − 6x + 15 + 240/x At x = 5: AC = 25 − 30 + 15 + 48 = 58 MC at x = 5 = 30 Since MC (30) < AC (58), average cost is still decreasing at x = 5. Minimum AC occurs when MC = AC: 3x^2 − 12x + 15 = x^2 − 6x + 15 + 240/x This is consistent with the general principle that MC = AC at minimum AC.

3. Elasticity of Demand — Worked Numerical Example

The demand function for a product is Q = 200 − 4P, where Q is quantity demanded and P is price.

Find: (i) Elasticity of demand at P = 20, (ii) Elasticity of demand at P = 30, (iii) Price at which demand is unitary elastic, (iv) Total revenue at P = 20 and P = 30.

Given: Q = 200 − 4P dQ/dP = −4 (i) Elasticity of Demand at P = 20: Q = 200 − 4(20) = 200 − 80 = 120 Ed = −(dQ/dP) × (P/Q) = −(−4) × (20/120) = 4 × (1/6) = 2/3 ≈ 0.67 Since Ed = 0.67 < 1, demand is INELASTIC at P = 20. A 1% increase in price leads to only 0.67% decrease in quantity. (ii) Elasticity of Demand at P = 30: Q = 200 − 4(30) = 200 − 120 = 80 Ed = −(dQ/dP) × (P/Q) = −(−4) × (30/80) = 4 × 0.375 = 1.5 Since Ed = 1.5 > 1, demand is ELASTIC at P = 30. A 1% increase in price leads to 1.5% decrease in quantity. (iii) Price at which demand is unitary elastic (Ed = 1): Ed = −(dQ/dP) × (P/Q) = 1 4 × P / (200 − 4P) = 1 4P = 200 − 4P 8P = 200 P = 25 At P = 25: Q = 200 − 100 = 100 Verify: Ed = 4 × (25/100) = 1 ✓ Demand is unitary elastic at P = ₹25. (iv) Total Revenue: TR = P × Q = P × (200 − 4P) = 200P − 4P^2 At P = 20: TR = 200(20) − 4(400) = 4000 − 1600 = ₹2400 At P = 30: TR = 200(30) − 4(900) = 6000 − 3600 = ₹2400 At P = 25 (unitary elastic): TR = 200(25) − 4(625) = 5000 − 2500 = ₹2500 Maximum TR occurs where Ed = 1 (unitary elasticity). This confirms the economic principle: total revenue is maximised when elasticity of demand equals 1.

Key economic insight:

When Ed < 1 (inelastic): Price ↑ → TR ↑ (P = 20 case) When Ed > 1 (elastic): Price ↑ → TR ↓ (P = 30 case) When Ed = 1 (unitary): TR is at its maximum Business implication: - If current demand is inelastic, raising price increases revenue. - If current demand is elastic, lowering price increases revenue.

Common Mistakes to Avoid

1. Forgetting the negative sign in elasticity of demand. The formula is Ed = −(dQ/dP)(P/Q). Since dQ/dP is typically negative (demand curves slope downward), the negative sign ensures Ed is positive. Dropping it gives negative elasticity, which is wrong.

2. Confusing maximum and minimum in the second derivative test. Remember: f''(c) < 0 means maximum (concave down, like an inverted bowl), f''(c) > 0 means minimum (concave up, like an upright bowl).

3. Not discarding negative values of x. In business problems, x represents output or quantity, which must be non-negative. Always check that critical points make economic sense.

4. Mixing up marginal and average. Marginal cost is dC/dx (rate of change of total cost). Average cost is C(x)/x (cost per unit). They are not the same thing, though they are equal at the minimum of AC.

5. Using only MR = MC without checking the second-order condition. MR = MC gives a critical point, but it could be a minimum. Always verify MR < MC (i.e., P''(x) < 0) to confirm maximum profit.

6. Computing Ed incorrectly by using dP/dQ instead of dQ/dP. The formula requires dQ/dP, the derivative of quantity with respect to price. If you have Q as a function of P, differentiate directly. If P is given as a function of Q, use the inverse function rule: dQ/dP = 1/(dP/dQ).

7. Ignoring domain restrictions in related rates problems. When using related rates, ensure the values you compute stay within the domain of the function (e.g., radius > 0, volume > 0).

8. Setting average cost equal to marginal cost and solving, but forgetting to verify it is a minimum. The condition MC = AC gives a critical point. Check that it corresponds to a minimum (not maximum) of AC by confirming the second derivative of AC is positive.

Exam Checklist

  • [ ] I can identify whether a function is increasing or decreasing using f'(x).
  • [ ] I can find critical points by setting f'(x) = 0 and solve correctly.
  • [ ] I can apply the first derivative test to classify critical points as max or min.
  • [ ] I can apply the second derivative test (f''(c) > 0 → min, f''(c) < 0 → max).
  • [ ] I can define and compute marginal cost (MC = dC/dx), marginal revenue (MR = dR/dx), and marginal profit.
  • [ ] I can compute average cost (AC = C/x) and average revenue (AR = R/x).
  • [ ] I can prove and apply MC = AC at minimum average cost.
  • [ ] I can set up and solve profit maximisation problems using MR = MC with second-order condition.
  • [ ] I can compute elasticity of demand using Ed = −(dQ/dP)(P/Q) and classify demand as elastic, inelastic, or unitary.
  • [ ] I can solve related rates problems using the chain rule (dy/dt = dy/dx × dx/dt).

Frequently Asked Questions

Set f'(x) = 0 to find critical points. Use the second derivative test: if f''(x) > 0 at the critical point, it is a minimum; if f''(x) < 0, it is a maximum; if f''(x) = 0, the test is inconclusive.
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