Business Mathematics TEST FILE | Work on Copy | NEB 12 BM Derivative Questions for TEST

FEEN Weekly Test Series

Business Mathematics: Applications of Derivatives

Topic: Maxima, Minima, Revenue, Cost & Profit Optimization

Full Marks: 100 | Time: 3 Hours

"Calculus is the mathematics of change. Master it to master business."

---

Section A: Analysis of Functions (Long Questions)

Question 1

Consider the cubic function y = 2x³ - 9x² + 12x + 5.

1. Find the first derivative (dy/dx) and determine the stationary points.
2. Use the second derivative test to determine the nature of these points (Local Maximum or Local Minimum).
3. Find the point of inflection where the concavity changes.

💡 FEEN Hint: Stationary points are where f'(x) = 0. Inflection is where f''(x) = 0. Remember: f''(x) > 0 means Minimum (Valley), f''(x) < 0 means Maximum (Hill).

Question 2

Examine the function f(x) = x³ - 6x² + 9x - 2 for its monotonic behavior.

1. Find the intervals in which the function is strictly increasing.
2. Find the intervals in which the function is strictly decreasing.
3. Sketch a rough graph based on the maxima, minima, and y-intercept.

💡 FEEN Hint: Factor the derivative into (x-a)(x-b). Use a number line test. If f'(x) is positive, graph goes up. If negative, graph goes down.

Question 3

Show that the function f(x) = x³ + 3x + 10 has neither a maximum nor a minimum value. Explain why this function is always increasing.

💡 FEEN Hint: Look at the discriminant of the quadratic derivative or simply notice if f'(x) is always positive. If the slope never hits zero, it never turns around!

Question 4

Find the absolute maximum and absolute minimum values of the function f(x) = 2x³ - 3x² - 12x on the closed interval [0, 3].

💡 FEEN Hint: Don't just check the turning points! You MUST also calculate the value of y at the endpoints (x=0 and x=3).

Question 5

For the function y = x³ - 3x² + 3x - 5, prove that the stationary point is actually a point of inflection and not a maximum or minimum.

💡 FEEN Hint: This happens when f'(x) = 0 AND f''(x) = 0 at the same spot. The curve flattens out but continues in the same direction.

Section B: Elasticity of Demand

Question 6

The demand function for a FEEN textbook is given by P = 100 - Q². Calculate the price elasticity of demand when the price is Rs. 84.

💡 FEEN Hint: First, find Q when P=84. Then find dP/dQ. Use formula: Ed = -(P/Q) * (1 / (dP/dQ)).

Question 7

Find the value of output (Q) for which the elasticity of demand is unity (unitary elastic) for the demand function P = 200 - 5Q.

💡 FEEN Hint: Set |Ed| = 1. Remember for linear demand P = a - bQ, unitary elasticity is usually at the midpoint of the demand curve.

Question 8

Prove that for a demand law of the form P * Q = K (Rectangular Hyperbola), the elasticity of demand is always equal to 1, regardless of the price.

💡 FEEN Hint: Express P as K/Q. Find dP/dQ. Substitute everything into the elasticity formula. Everything should cancel out!

Question 9

Given the demand function Q = 50 - 2P, calculate the elasticity of demand at P = 10 and P = 20. Interpret which price point is more sensitive to changes.

💡 FEEN Hint: A higher elasticity value (Ed > 1) means customers are very sensitive to price changes (Luxury goods).

Question 10

Derive the relationship: Marginal Revenue (MR) = Average Revenue (AR) * (1 - 1/Ed).

💡 FEEN Hint: Start with R = P*Q. Differentiate R w.r.t Q using the product rule. Factor out P (which is AR).

Section C: Revenue, Cost & Profit Optimization

Question 11

A manufacturer has a total cost function C(Q) = 2Q² + 10Q + 500 and a demand function P = 50 - 2Q.

1. Formulate the Total Revenue (R) and Total Profit (π) functions.
2. Find the level of output (Q) that maximizes Total Profit.
3. Calculate the Maximum Profit amount.

💡 FEEN Hint: Revenue = P * Q. Profit = Revenue - Cost. Differentiate Profit and set to 0. Check if 2nd derivative is negative.

Question 12

Given the Total Cost function C = 1/3 Q³ - 2Q² + 10Q + 20, find the output level at which the Marginal Cost (MC) is minimum.

💡 FEEN Hint: First find MC (derivative of C). Then, find the derivative of MC and set THAT to zero to minimize it.

Question 13

The cost function is given by C = Q² + 4Q + 100. Find the output level where Average Cost (AC) is minimum. Verify that at this point, AC = MC.

💡 FEEN Hint: AC = C/Q. Differentiate AC w.r.t Q and set to 0. Plug that Q into both AC and MC equations to show they are equal.

Question 14 (Market Equilibrium)

In a perfect competition market, the price is fixed at P = Rs. 20. The firm's total cost function is C = 0.5Q² + 5Q + 100. Find the profit-maximizing output.

💡 FEEN Hint: In perfect competition, P = MR. So simply set P = MC and solve for Q.

Question 15

A monopolist has a demand curve Q = 100 - P and total costs C = 20 + 2Q. Determine the price and quantity that maximizes Revenue.

💡 FEEN Hint: Careful! Maximize REVENUE, not Profit. Convert demand to P = 100 - Q. R = (100-Q)Q. Set dR/dQ = 0.

Section D: Applied Word Problems

Question 16

Divide the number 120 into two parts such that the sum of their squares is minimum.

💡 FEEN Hint: Let parts be x and (120-x). Function S = x² + (120-x)². Minimize S.

Question 17

A rectangular garden is to be fenced on three sides with 100m of fencing, the fourth side being a straight wall. Find the dimensions that maximize the area.

💡 FEEN Hint: Perimeter Constraint: 2x + y = 100 (only 3 sides). Area A = xy. Substitute y = 100 - 2x into Area equation.

Question 18

The total cost of producing x radio sets per day is Rs. (x²/4 + 35x + 25) and the price per set at which they may be sold is Rs. (50 - x/2). Find the daily output to maximize total profit.

💡 FEEN Hint: Remember: Profit = (Price × Quantity) - Cost. Watch your fractions when differentiating!

Question 19

Find two positive numbers whose product is 64 and their sum is minimum.

💡 FEEN Hint: xy = 64, so y = 64/x. Minimize S = x + 64/x. This is a classic derivative problem.

Question 20 (The Challenge)

A tour operator charges Rs. 136 per passenger for 100 passengers. For every additional passenger, the price per person decreases by Rs. 0.40. Find the number of passengers that maximizes the total revenue.

💡 FEEN Hint: Let x be additional passengers. Quantity = 100+x. Price = 136 - 0.40x. Revenue = (100+x)(136-0.40x). Maximize this quadratic.

🚀 FEEN Exam Strategy for Derivatives

• The "Zero" Rule: For any Max/Min/Stationary point problem, your first step is ALWAYS finding dy/dx and setting it to 0.
• The "Double Check": Always use the second derivative (d²y/dx²) to confirm if your answer is a Max (negative) or Min (positive).
• Units Matter: In business math, check if Price is in Rs. or Hundreds. Check if Q is units or thousands.
• Profit vs Revenue: Don't confuse them! Revenue is just money IN. Profit is Revenue minus Cost.

"Practice makes permanent. Good luck from Team FEEN!"