Derivatives in Business: Cost, Revenue, Profit & Optimization

TAKEAWAYS:

• Derivatives measure instantaneous rates of change (e.g., how cost changes with production).
• Critical points (where $f^{\prime }(x)=0$) help find maxima/minima for profit or cost.
• Second derivative tests confirm whether a critical point is a profit peak or cost valley.
• Businesses use derivatives to optimize pricing, production, and resource allocation.
• Marginal cost/revenue (first derivative) shows the cost/revenue of one extra unit.
• Elasticity of demand (using derivatives) helps predict sales changes from price adjustments.

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1. Why Derivatives Matter in Business

Businesses need to make smart decisions to earn more profit and reduce costs. Derivatives help by showing:

• How cost, revenue, and profit change when production or sales change.
• The best price to set for maximum profit.
• The optimal quantity to produce for minimum cost.

Example: A shopkeeper wants to know:

• How much extra cost does producing 1 more shirt add?
• At what price should they sell to maximize profit?

Derivatives answer these questions!

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2. Cost, Revenue, and Profit Functions

Before using derivatives, we need these three functions:

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3. Marginal Cost and Marginal Revenue

• Marginal Cost (MC): The cost of producing one extra unit. $$MC=C^{\prime }(x)=\frac{d}{dx}[C(x)]$$
• Marginal Revenue (MR): The extra revenue from selling one extra unit. $$MR=R^{\prime }(x)=\frac{d}{dx}[R(x)]$$

Worked Example 1: Finding Marginal Cost & Revenue Suppose:

• Cost function: $C(x)=0.5x^2+10x+100$
• Revenue function: $R(x)=50x-0.5x^2$

Step 1: Find $C^{\prime }(x)$ and $R^{\prime }(x)$. $$C^{\prime }(x)=\frac{d}{dx}(0.5x^2+10x+100)=x+10$$ $$R^{\prime }(x)=\frac{d}{dx}(50x-0.5x^2)=50-x$$

Step 2: Interpret at $x=20$.

• Marginal Cost at 20 units: $C^{\prime }(20)=20+10=20$ (cost of 21st unit = ₹20)
• Marginal Revenue at 20 units: $R^{\prime }(20)=50-20=30$ (extra income from 21st unit = ₹30)

Conclusion:

• If the shop sells the 21st unit, they gain ₹10 (₹30 revenue - ₹20 cost).
• If they sell more than 21 units, marginal revenue drops below marginal cost (loss starts).

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4. Finding Maximum Profit

Profit is maximized where: $$P^{\prime }(x)=0\text{(First Derivative Test)}$$ And confirmed by: $$P^{\prime \prime }(x)<0\text{(Second Derivative Test: concave down = maximum)}$$

Worked Example 2: Maximizing Profit Given: $$P(x)=-2x^2+100x-200$$

Step 1: Find $P^{\prime }(x)$. $$P^{\prime }(x)=\frac{d}{dx}(-2x^2+100x-200)=-4x+100$$

Step 2: Set $P^{\prime }(x)=0$ to find critical points. $$-4x+100=0⟹x=25$$

Step 3: Use the Second Derivative Test. $$P^{\prime \prime }(x)=\frac{d}{dx}(-4x+100)=-4$$ Since $P^{\prime \prime }(x)=-4<0$, $x=25$ gives maximum profit.

Step 4: Find maximum profit. $$P(25)=-2(25{)}^2+100(25)-200=-1250+2500-200=1050$$

Conclusion:

• Optimal production = 25 units
• Maximum profit = ₹1050

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5. Finding Minimum Cost

Cost is minimized where: $$C^{\prime }(x)=0\text{and}{C}^{\prime \prime }(x)>0$$

Worked Example 3: Minimizing Cost Given: $$C(x)=0.1x^3-6x^2+100x+500$$

Step 1: Find $C^{\prime }(x)$. $$C^{\prime }(x)=0.3x^2-12x+100$$

Step 2: Set $C^{\prime }(x)=0$. $$0.3x^2-12x+100=0$$ Solve using the quadratic formula: $$x=\frac{12\pm \sqrt{(-12{)}^2-4(0.3)(100)}}{2(0.3)}=\frac{12\pm \sqrt{144-120}}{0.6}=\frac{12\pm 5.3}{0.6}$$ $$x_1=\frac{12+5.3}{0.6}\approx 29\text{(Not feasible, as cost rises after this)}$$ $$x_2=\frac{12-5.3}{0.6}\approx 11.17$$

Step 3: Check $C^{\prime \prime }(x)$. $$C^{\prime \prime }(x)=0.6x-12$$ At $x=11.17$: $$C^{\prime \prime }(11.17)=0.6(11.17)-12\approx 6.7-12=-5.3\text{(Not > 0!)}$$ Mistake: The second derivative is negative, so this is a maximum, not a minimum.

Correction: We need to check behavior or use First Derivative Test.

• For $x<11.17$, $C^{\prime }(x)$ is positive (cost increasing).
• For $11.17<x<29$, $C^{\prime }(x)$ is negative (cost decreasing).
• For $x>29$, $C^{\prime }(x)$ is positive (cost increasing again).

Conclusion:

• Minimum cost occurs at $x=29$ (where cost stops decreasing and starts increasing).

Step 4: Find minimum cost. $$C(29)=0.1(29{)}^3-6(29{)}^2+100(29)+500\approx 2438.9-5046+2900+500=792.9$$

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6. Price Elasticity of Demand (PED)

Measures how sensitive demand is to price changes. $$\text{PED}=\frac{dQ}{dP}\times \frac{P}{Q}$$

• If $|\text{PED}|>1$: Elastic (demand changes a lot with price).
• If $|\text{PED}|<1$: Inelastic (demand changes little with price).

Worked Example 4: Calculating PED Given demand function: $$Q=100-2P$$ Find PED at $P=20$.

Step 1: Find $\frac{dQ}{dP}$. $$\frac{dQ}{dP}=-2$$

Step 2: Find $Q$ at $P=20$. $$Q=100-2(20)=60$$

Step 3: Calculate PED. $$\text{PED}=(-2)\times \frac{20}{60}=-\frac{40}{60}=-\frac{2}{3}\approx -0.67$$

Interpretation:

• $|\text{PED}|=0.67<1$ → Inelastic demand.
• If the shop increases price, revenue will increase (since demand doesn’t drop much).

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7. Applications in Business

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8. Common Mistakes to Avoid

1. Ignoring the Second Derivative Test
   • Just setting $f^{\prime }(x)=0$ is not enough! Always check $f^{\prime \prime }(x)$ to confirm max/min.
2. Misinterpreting Marginal Values
   • Marginal cost/revenue is not the average cost/revenue.
3. Forgetting Units
   • Always label answers (e.g., "₹ per unit" for marginal cost).
4. Skipping Feasibility Checks
   • If $x=-5$ is a critical point but production can’t be negative, discard it.

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Exam Tip: How to Score Full Marks

1. Show All Steps

   • NEB examiners deduct marks if steps are skipped.
   • Example:
     • Given $P(x)=-x^2+10x$, find max profit.
     • Wrong: $P^{\prime }(x)=10$ → $x=5$ (❌ No derivative calculated!)
     • Right: $$P^{\prime }(x)=-2x+10=0⟹x=5$$ $$P^{\prime \prime }(x)=-2<0⟹\text{Maximum at }x=5$$

2. Use Real-World Context

   • Always interpret answers in business terms.
   • Example:
     • "The company should produce 50 units to maximize profit of ₹2000."

3. Practice Graph Sketching

   • NEB often asks to sketch cost/revenue/profit curves and mark critical points.
   • Example Question:

     "Given $C(x)=x^2+10x$ and $R(x)=30x$, sketch the profit function and find the break-even points."

4. Memorize Key Formulas

   Concept	Formula
   Marginal Cost	$C^{\prime }(x)$
   Marginal Revenue	$R^{\prime }(x)$
   Profit Maximization	$P^{\prime }(x)=0$ and $P^{\prime \prime }(x)<0$
   Cost Minimization	$C^{\prime }(x)=0$ and $C^{\prime \prime }(x)>0$
   Price Elasticity of Demand	$\text{PED}=\frac{dQ}{dP}\times \frac{P}{Q}$

5. Watch for Tricks in Questions

   • Sometimes, the second derivative is zero (test fails). Use the First Derivative Test instead.
   • Example:

     "If $f^{\prime }(x)=(x-2{)}^3$, find critical points and classify them."

     • Solution: $f^{\prime }(x)=0$ at $x=2$, but $f^{\prime \prime }(x)=0$. Check signs of $f^{\prime }(x)$ around $x=2$:
       • For $x<2$, $f^{\prime }(x)<0$ (decreasing).
       • For $x>2$, $f^{\prime }(x)>0$ (increasing).
       • Conclusion: $x=2$ is a minimum.

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NEB-Style Practice Questions

Section A: Short Answer (2 marks each)

1. Define marginal cost and write its formula.
2. If $P(x)=5x-x^2$, find the production level for maximum profit.
3. What does $P^{\prime \prime }(x)<0$ indicate about the profit function?
4. Calculate the price elasticity of demand for $Q=50-3P$ at $P=5$.

Section B: Long Answer (5 marks each) 5. A company’s cost and revenue functions are: $$C(x)=0.5x^2+20x+100,R(x)=50x-0.2x^2$$

• Find the break-even points.
• Determine the production level for maximum profit.
• Calculate the maximum profit.

6. The demand function for a product is $Q=100-4P$.
   • Find the price elasticity of demand at $P=10$.
   • Should the company increase or decrease the price to increase total revenue? Justify.

Section C: Problem Solving (7 marks) 7. A factory’s profit function is $P(x)=-2x^3+30x^2-100x+50$.

• Find the critical points.
• Classify each critical point (max/min/inflection).
• Determine the production level for maximum profit and the profit value.

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Answers to Practice Questions

1. Marginal cost = Cost of producing one extra unit. Formula: $MC=C^{\prime }(x)$.
2. $P^{\prime }(x)=5-2x=0⟹x=2.5$.
3. Indicates a local maximum (profit peaks at that point).
4. $\frac{dQ}{dP}=-3$, $Q=50-3(5)=35$, $\text{PED}=-3\times \frac{5}{35}=-0.43$.
5. • Break-even: $R(x)=C(x)⟹50x-0.2x^2=0.5x^2+20x+100⟹0.7x^2-30x+100=0$. Solving: $x\approx 10$ or $x\approx 32.86$.
   • Max profit: $P^{\prime }(x)=R^{\prime }(x)-C^{\prime }(x)=50-0.4x-(x+20)=30-1.4x=0⟹x\approx 21.43$.
   • Max profit: $P(21.43)\approx ₹230.5$.
6. • $\text{PED}=-4\times \frac{10}{60}=-0.67$ (inelastic).
   • Since $|\text{PED}|<1$, increase price to increase revenue.
7. • Critical points: $P^{\prime }(x)=-6x^2+60x-100=0⟹x\approx 3.33$ or $x\approx 6.67$.
   • $P^{\prime \prime }(x)=-12x+60$:
     • At $x=3.33$, $P^{\prime \prime }(3.33)>0$ → Minimum.
     • At $x=6.67$, $P^{\prime \prime }(6.67)<0$ → Maximum.
   • Max profit at $x=6.67$, $P(6.67)\approx ₹120$.