OP Malhotra Class 12 Maths Solutions Chapter 26 Application of Calculus in Commerce and Economics Exercise 26 (D)

Question 1. The demand function for a manufacturer's product is \( p = 20 - \frac{x}{4} \), where x is units and p is the price per unit. At what value of x will there be revenue? What is the maximum revenue?
Answer: The demand function is given as \( p = 20 - \frac{x}{4} \), where p is the price per unit and x is the number of units. First, we find the revenue function, R(x), which is price multiplied by quantity: Revenue function \( R(x) = px = \left(20 - \frac{x}{4}\right)x = 20x - \frac{x^2}{4} \) To find the maximum revenue, we need to take the first derivative of R(x) and set it to zero. \( \frac{dR}{dx} = 20 - \frac{2x}{4} = 20 - \frac{x}{2} \) Set the first derivative to zero to find critical points: \( 20 - \frac{x}{2} = 0 \) \( 20 = \frac{x}{2} \) \( x = 40 \) Next, we use the second derivative to confirm if this value of x gives a maximum. \( \frac{d^2R}{dx^2} = -\frac{1}{2} \) Since the second derivative is \( -\frac{1}{2} \), which is less than 0, the revenue is maximum when \( x = 40 \) units. Now, we calculate the maximum revenue by substituting \( x = 40 \) into the revenue function: Maximum revenue \( = R(40) = 20(40) - \frac{(40)^2}{4} \) \( = 800 - \frac{1600}{4} \) \( = 800 - 400 \) \( = 400 \) Therefore, the maximum revenue is 400. The maximum revenue is Rs. 400 when 40 units are produced.In simple words: First, we write down the total money we can get from selling things (revenue). Then, we find the number of items that will give us the most money by using a math trick called "derivatives". We check to make sure it's the highest point, not the lowest. Finally, we put that number back into our revenue formula to find the biggest possible income.

🎯 Exam Tip: Always remember that revenue is quantity multiplied by price. For maximization problems, the second derivative test is crucial to distinguish between maximum and minimum points.

Question 2. The unit demand function is \( x = \frac{1}{3}(24 - 2p) \), where x is the number of units demanded and p is the price per unit. Find (i) the revenue function R in terms of price p. (ii) the price and number of units demanded for which the revenue is maximum.
Answer: Given demand function: \( x = \frac{1}{3}(24 - 2p) \) (i) To find the revenue function R in terms of price p: Revenue R is given by \( R(p) = px \). Substitute the expression for x: \( R(p) = p \left[\frac{1}{3}(24 - 2p)\right] \) \( R(p) = \frac{1}{3}(24p - 2p^2) \) (ii) To find the price and number of units for which revenue is maximum: We need to maximize \( R(p) \). Take the first derivative of \( R(p) \) with respect to p: \( \frac{dR}{dp} = \frac{1}{3}(24 - 4p) \) Set the first derivative to zero to find critical points: \( \frac{1}{3}(24 - 4p) = 0 \) \( 24 - 4p = 0 \) \( 4p = 24 \) \( p = 6 \) Next, we use the second derivative to confirm if this value of p gives a maximum. \( \frac{d^2R}{dp^2} = \frac{d}{dp}\left[\frac{1}{3}(24 - 4p)\right] = \frac{1}{3}(-4) = -\frac{4}{3} \) Since the second derivative is \( -\frac{4}{3} \), which is less than 0, the revenue is maximum when \( p = 6 \). Now, find the number of units x when \( p = 6 \): \( x = \frac{1}{3}(24 - 2p) = \frac{1}{3}(24 - 2 \times 6) \) \( x = \frac{1}{3}(24 - 12) = \frac{1}{3}(12) \) \( x = 4 \) So, the revenue is maximum when the price is Rs. 6 per unit and 4 units are demanded.In simple words: First, we create a formula for how much money we get (revenue) based on the price. Then, we use special math steps to find the exact price that will give us the most money. Once we have that price, we figure out how many items will be sold at that price. This helps find the "sweet spot" for making the most money.

🎯 Exam Tip: Always ensure you differentiate with respect to the correct variable (p or x) as specified in the question or for the function you are maximizing.

Question 3. The demand function for a particular commodity is \( p = 15e^{-x/3} \) for \( 0 \leq x \leq 8 \), where p is the price per unit and x is the number of units demanded. Determine the price and quantity for which total revenue is maximum.
Answer: The demand function is given as \( p = 15e^{-x/3} \), where p is the price per unit and x is the number of units demanded. The total revenue function R(x) is price multiplied by quantity: \( R(x) = px = 15xe^{-x/3} \) To find the maximum revenue, we need to take the first derivative of R(x) with respect to x. We use the product rule \( (uv)' = u'v + uv' \), where \( u = 15x \) and \( v = e^{-x/3} \). \( u' = 15 \) \( v' = e^{-x/3} \left(-\frac{1}{3}\right) = -\frac{1}{3}e^{-x/3} \) So, \( \frac{dR}{dx} = 15e^{-x/3} + 15x \left(-\frac{1}{3}e^{-x/3}\right) \) \( \frac{dR}{dx} = 15e^{-x/3} - 5xe^{-x/3} \) Factor out \( 5e^{-x/3} \): \( \frac{dR}{dx} = 5e^{-x/3}(3 - x) \) Set the first derivative to zero to find critical points: \( 5e^{-x/3}(3 - x) = 0 \) Since \( 5e^{-x/3} \) is never zero, we must have \( 3 - x = 0 \), which means \( x = 3 \). Next, we use the second derivative to confirm if this value of x gives a maximum. We differentiate \( \frac{dR}{dx} = 15e^{-x/3} - 5xe^{-x/3} \) again. For \( 15e^{-x/3} \), the derivative is \( 15e^{-x/3}(-\frac{1}{3}) = -5e^{-x/3} \). For \( -5xe^{-x/3} \), use the product rule with \( u = -5x \) and \( v = e^{-x/3} \). \( u' = -5 \) \( v' = -\frac{1}{3}e^{-x/3} \) So, the derivative of \( -5xe^{-x/3} \) is \( (-5)e^{-x/3} + (-5x)(-\frac{1}{3}e^{-x/3}) = -5e^{-x/3} + \frac{5x}{3}e^{-x/3} \) Combine these to get the second derivative: \( \frac{d^2R}{dx^2} = -5e^{-x/3} + (-5e^{-x/3} + \frac{5x}{3}e^{-x/3}) \) \( \frac{d^2R}{dx^2} = -10e^{-x/3} + \frac{5x}{3}e^{-x/3} = e^{-x/3}\left(\frac{5x}{3} - 10\right) \) Now, evaluate the second derivative at \( x = 3 \): \( \frac{d^2R}{dx^2}\Big|_{x=3} = e^{-3/3}\left(\frac{5(3)}{3} - 10\right) = e^{-1}(5 - 10) = -5e^{-1} = -\frac{5}{e} \) Since \( -\frac{5}{e} \) is less than 0, the revenue is maximum when \( x = 3 \) units. Finally, find the price p when \( x = 3 \): \( p = 15e^{-x/3} = 15e^{-3/3} = 15e^{-1} = \frac{15}{e} \) Therefore, the total revenue is maximum when the quantity demanded is 3 units and the price per unit is \( \frac{15}{e} \). Exponential functions are common in economic models, especially for demand or growth over time.In simple words: We are given how the price changes with the number of items sold. We write a formula for total money earned. Then, we use math to find the exact number of items that will make the most money. After that, we find the price for that number of items to get the highest profit.

🎯 Exam Tip: Remember to apply the product rule correctly when differentiating functions involving both x and an exponential of x, and always evaluate the second derivative to confirm a maximum.

Question 4. A club has 1000 members who are each paying Rs. 100 per month. The club proposes to increase the monthly membership fee and it is expected that for every increase of Rs. 1, five members will discontinue the service. Find what increase will yield maximum revenue and what will this revenue be?
Answer: Let \( x \) be the increase in the monthly fee (in Rs.). The initial number of members is 1000, and the initial fee is Rs. 100 per month. For every Rs. 1 increase in fee, 5 members discontinue. So, for an \( x \) Rs. increase, \( 5x \) members will discontinue. New number of members = \( 1000 - 5x \) New fee per member = \( 100 + x \) The total revenue function R(x) is the new fee multiplied by the new number of members: \( R(x) = (100 + x)(1000 - 5x) \) Expand the expression: \( R(x) = 100 \times 1000 + 100 \times (-5x) + x \times 1000 + x \times (-5x) \) \( R(x) = 100000 - 500x + 1000x - 5x^2 \) \( R(x) = -5x^2 + 500x + 100000 \) To find the maximum revenue, take the first derivative of R(x) with respect to x: \( \frac{dR}{dx} = -10x + 500 \) Set the first derivative to zero to find critical points: \( -10x + 500 = 0 \) \( 10x = 500 \) \( x = 50 \) Next, use the second derivative to confirm if this value of x gives a maximum. \( \frac{d^2R}{dx^2} = -10 \) Since the second derivative is \( -10 \), which is less than 0, the revenue is maximum when \( x = 50 \). This means an increase of Rs. 50 in the monthly fee will yield maximum revenue. Now, calculate the maximum revenue at \( x = 50 \): Maximum Revenue \( = R(50) = (100 + 50)(1000 - 5 \times 50) \) \( = (150)(1000 - 250) \) \( = 150 \times 750 \) \( = 112500 \) The maximum revenue will be Rs. 1,12,500. Understanding how price changes affect both the price per unit and the quantity sold is key in economics.In simple words: We figure out how a price increase changes both the cost for each person and the total number of people. We then make a formula for the total money the club gets. Using math, we find the best amount to increase the fee to make the most money, and then we calculate that highest amount of money.

🎯 Exam Tip: Carefully set up the revenue function as a product of the new price and new quantity, ensuring all changes (like member discontinuation) are correctly accounted for.

Question 5. A company charges Rs. 550 for a transistor set on orders of 50 or less sets. The charge is reduced by Rs. 5 per set for each set ordered in excess of 50. Find the largest size order company should allow so as to receive maximum revenue.
Answer: Let \( x \) be the number of sets ordered in excess of 50. The total number of sets in an order will be \( 50 + x \). The charge is reduced by Rs. 5 for each set ordered in excess of 50. So, the total reduction in price per set is \( 5x \). New price per set = \( 550 - 5x \) The total revenue function R(x) is the total number of sets multiplied by the new price per set: \( R(x) = (50 + x)(550 - 5x) \) Expand the expression: \( R(x) = 50 \times 550 + 50 \times (-5x) + x \times 550 + x \times (-5x) \) \( R(x) = 27500 - 250x + 550x - 5x^2 \) \( R(x) = -5x^2 + 300x + 27500 \) To find the maximum revenue, take the first derivative of R(x) with respect to x: \( \frac{dR}{dx} = -10x + 300 \) Set the first derivative to zero to find critical points: \( -10x + 300 = 0 \) \( 10x = 300 \) \( x = 30 \) Next, use the second derivative to confirm if this value of x gives a maximum. \( \frac{d^2R}{dx^2} = -10 \) Since the second derivative is \( -10 \), which is less than 0, the revenue is maximum when \( x = 30 \). The value \( x=30 \) represents the number of sets in excess of 50. So, the largest size order the company should allow for maximum revenue is: Total sets = \( 50 + x = 50 + 30 = 80 \) sets. Many businesses use tiered pricing to encourage larger orders, but there's an optimal point for maximizing income.In simple words: We define how many extra items are ordered. Then, we create a formula for the total money made, based on the price changing for larger orders. We use math to find the exact number of extra items that brings in the most money, and then we add that to the original number of items to get the total order size.

🎯 Exam Tip: Be careful when defining 'x' – make sure it represents the variable quantity (in this case, the excess sets) that influences both the price and total units.

Question 6. A steel plant is capable of producing x tonnes per day of a low grade steel and y tonnes per day of a high grade steel, where \( y = \frac{42-5x}{10-x} \). If the fixed market price of low grade steel is half of high grade steel, show that 6 tonnes of low grade steel are produced per day for maximum total revenue.
Answer: Let \( x \) be the quantity (in tonnes) of low grade steel produced per day. Let \( y \) be the quantity (in tonnes) of high grade steel produced per day. Given relation: \( y = \frac{42-5x}{10-x} \) Let the fixed market price of high grade steel be \( P_H \). Then the fixed market price of low grade steel is \( P_L = \frac{1}{2} P_H \). The total revenue \( TR(x) \) is the sum of revenue from low grade steel and high grade steel: \( TR(x) = x \cdot P_L + y \cdot P_H \) \( TR(x) = x \cdot \frac{P_H}{2} + \left(\frac{42-5x}{10-x}\right) \cdot P_H \) Since \( P_H \) is a positive constant, maximizing \( TR(x) \) is equivalent to maximizing the function \( f(x) = \frac{x}{2} + \frac{42-5x}{10-x} \). Take the first derivative of \( f(x) \) with respect to x: \( f'(x) = \frac{d}{dx}\left(\frac{x}{2}\right) + \frac{d}{dx}\left(\frac{42-5x}{10-x}\right) \) For the second term, use the quotient rule \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \), where \( u = 42-5x \) and \( v = 10-x \). \( u' = -5 \) \( v' = -1 \) So, \( \frac{d}{dx}\left(\frac{42-5x}{10-x}\right) = \frac{(-5)(10-x) - (42-5x)(-1)}{(10-x)^2} \) \( = \frac{-50 + 5x + 42 - 5x}{(10-x)^2} = \frac{-8}{(10-x)^2} \) Now, combine the derivatives: \( f'(x) = \frac{1}{2} - \frac{8}{(10-x)^2} \) Set the first derivative to zero to find critical points: \( \frac{1}{2} - \frac{8}{(10-x)^2} = 0 \) \( \frac{1}{2} = \frac{8}{(10-x)^2} \) \( (10-x)^2 = 16 \) Take the square root of both sides: \( 10-x = \pm 4 \) Case 1: \( 10-x = 4 \Rightarrow x = 6 \) Case 2: \( 10-x = -4 \Rightarrow x = 14 \) Given the expression for \( y \), the denominator \( (10-x) \) implies \( x \neq 10 \). Also, for \( y \) to be a positive quantity of steel, considering \( 42-5x \) is positive for \( x < 8.4 \), then \( 10-x \) should also be positive, meaning \( x < 10 \). Thus, \( x=6 \) is the relevant value. Next, use the second derivative to confirm if \( x=6 \) gives a maximum. \( f''(x) = \frac{d}{dx}\left(\frac{1}{2} - 8(10-x)^{-2}\right) \) \( f''(x) = 0 - 8(-2)(10-x)^{-3}(-1) = -16(10-x)^{-3} = -\frac{16}{(10-x)^3} \) Evaluate \( f''(x) \) at \( x = 6 \): \( f''(6) = -\frac{16}{(10-6)^3} = -\frac{16}{4^3} = -\frac{16}{64} = -\frac{1}{4} \) Since \( f''(6) = -\frac{1}{4} \), which is less than 0, total revenue is maximum when \( x = 6 \) tonnes of low grade steel are produced per day. The second derivative test helps confirm if a critical point is a maximum or minimum, which is essential for optimization problems.In simple words: We write a formula for the total money earned from selling both kinds of steel, considering their prices and quantities. Then, we use special math rules to find the exact amount of low-grade steel that should be made to get the most money. We also check our answer to make sure it's truly the highest point of money earned.

🎯 Exam Tip: For problems involving related production quantities, ensure the total revenue function correctly combines the revenue from all products based on their individual prices and quantities before differentiation.

Question 7. A tour operator engages a bus of seating capacity of 50 for taking children on an excursion and charges Rs. 100 per child with an additional charge of Rs. 2.50 for each subsequent cancellation. Determine the total revenue R(x), as a function of x, the number of cancellation received prior to the departure date. What is the value of x for which R(x) maximum.
Answer: Let \( x \) be the number of cancellations received. The bus has a seating capacity of 50 children. Number of children attending the excursion = \( 50 - x \) The original charge per child is Rs. 100. There's an additional charge of Rs. 2.50 for each cancellation. So, the total additional charge added to each child's ticket due to cancellations is \( 2.50x \). New charge per child = \( 100 + 2.5x \) The total revenue function R(x) is the number of children attending multiplied by the new charge per child: \( R(x) = (50 - x)(100 + 2.5x) \) Expand the expression: \( R(x) = 50 \times 100 + 50 \times 2.5x - x \times 100 - x \times 2.5x \) \( R(x) = 5000 + 125x - 100x - 2.5x^2 \) \( R(x) = -2.5x^2 + 25x + 5000 \) To find the maximum revenue, take the first derivative of R(x) with respect to x: \( \frac{dR}{dx} = -5x + 25 \) Set the first derivative to zero to find critical points: \( -5x + 25 = 0 \) \( 5x = 25 \) \( x = 5 \) Next, use the second derivative to confirm if this value of x gives a maximum. \( \frac{d^2R}{dx^2} = -5 \) Since the second derivative is \( -5 \), which is less than 0, the revenue is maximum when \( x = 5 \). So, the total revenue R(x) is maximum when there are 5 cancellations. This scenario demonstrates that sometimes adjusting prices based on demand, even for cancellations, can lead to higher overall profits.In simple words: We first set up a formula for the total money earned, considering how many children attend and how the price changes with cancellations. Then, we use special math steps to find the exact number of cancellations that will make the most money for the tour operator.

🎯 Exam Tip: Carefully read whether the extra charge applies per cancellation or per remaining passenger. Here it implies the total additional charge is distributed among the remaining passengers.

Question 8. The revenue function of a product is given by the relation \( y = 4000000 - (x - 2000)^2 \), where y is the total revenue and x is the number of units sold. Find (i) the number of units sold which maximizes total revenue, (ii) the amount of maximum revenue.
Answer: The given total revenue function is \( y = R(x) = 4000000 - (x - 2000)^2 \). (i) To find the number of units sold which maximizes total revenue: Take the first derivative of R(x) with respect to x: \( \frac{dR}{dx} = -2(x - 2000)(1) = -2(x - 2000) \) Set the first derivative to zero to find critical points: \( -2(x - 2000) = 0 \) \( x - 2000 = 0 \) \( x = 2000 \) Next, use the second derivative to confirm if this value of x gives a maximum. \( \frac{d^2R}{dx^2} = -2 \) Since the second derivative is \( -2 \), which is less than 0, the revenue is maximum when \( x = 2000 \). So, the number of units sold which maximizes total revenue is 2000 units. (ii) To find the amount of maximum revenue: Substitute \( x = 2000 \) into the revenue function: Maximum Revenue \( = R(2000) = 4000000 - (2000 - 2000)^2 \) \( = 4000000 - (0)^2 \) \( = 4000000 \) The amount of maximum revenue is Rs. 40,00,000. This revenue function is a parabola opening downwards, so its vertex represents the maximum revenue.In simple words: We have a formula for how much money is made when different numbers of items are sold. We use math to find the exact number of items that will make the most money. Then, we put that number back into the formula to calculate what that highest amount of money will be.

🎯 Exam Tip: The derivative of \( (ax+b)^n \) is \( n(ax+b)^{n-1} \cdot a \). Remember to apply the chain rule correctly for derivatives like \( (x-2000)^2 \).

Question 9. The total cost and demand functions of an item are given by \( C(x) = \frac{x^3}{3} - 7x^2 + 111x + 50 \) and \( p = 100 - x \) respectively. Write the total revenue function and the profit function. Find the profit maximizing level of output x and the maximum profit.
Answer: Given cost function: \( C(x) = \frac{x^3}{3} - 7x^2 + 111x + 50 \) Given demand function: \( p = 100 - x \) The total revenue function R(x) is price multiplied by quantity: \( R(x) = px = (100 - x)x = 100x - x^2 \) The profit function P(x) is total revenue minus total cost: \( P(x) = R(x) - C(x) \) \( P(x) = (100x - x^2) - \left(\frac{x^3}{3} - 7x^2 + 111x + 50\right) \) \( P(x) = 100x - x^2 - \frac{x^3}{3} + 7x^2 - 111x - 50 \) Combine like terms: \( P(x) = -\frac{x^3}{3} + 6x^2 - 11x - 50 \) To find the profit maximizing level of output, take the first derivative of P(x) with respect to x: \( \frac{dP}{dx} = -\frac{3x^2}{3} + 6(2x) - 11 = -x^2 + 12x - 11 \) Set the first derivative to zero to find critical points: \( -x^2 + 12x - 11 = 0 \) Multiply by -1: \( x^2 - 12x + 11 = 0 \) Factorize the quadratic equation: \( (x - 1)(x - 11) = 0 \) So, \( x = 1 \) or \( x = 11 \). Next, use the second derivative to confirm which value of x gives a maximum. \( \frac{d^2P}{dx^2} = -2x + 12 \) At \( x = 1 \): \( \frac{d^2P}{dx^2}\Big|_{x=1} = -2(1) + 12 = 10 \) (This is greater than 0, so it's a minimum). At \( x = 11 \): \( \frac{d^2P}{dx^2}\Big|_{x=11} = -2(11) + 12 = -22 + 12 = -10 \) (This is less than 0, so it's a maximum). The profit maximizing level of output is \( x = 11 \) units. Now, calculate the maximum profit at \( x = 11 \): Maximum Profit \( = P(11) = -\frac{(11)^3}{3} + 6(11)^2 - 11(11) - 50 \) \( P(11) = -\frac{1331}{3} + 6(121) - 121 - 50 \) \( P(11) = -\frac{1331}{3} + 726 - 121 - 50 \) \( P(11) = -\frac{1331}{3} + 555 \) To add these, find a common denominator: \( P(11) = \frac{-1331 + 555 \times 3}{3} = \frac{-1331 + 1665}{3} = \frac{334}{3} \) The maximum profit is Rs. \( \frac{334}{3} \). Profit maximization is a key goal for businesses, balancing production costs and market prices.In simple words: First, we define how much money we get (revenue) and how much money we spend (cost). We subtract cost from revenue to get the profit. Then, we use math rules to find the number of items we should make to get the most profit. Finally, we calculate what that biggest profit amount will be.

🎯 Exam Tip: Always check both critical points with the second derivative to correctly identify maximums and minimums, especially for cubic profit functions.

Question 10. A company has produced x items and the total cost C and total revenue R are given by the equation \( C = 100 + 0.015x^2 \) and \( R = 3x \). Find how many items should be produced to maximize the profit. What is this profit?
Answer: Given cost function: \( C(x) = 100 + 0.015x^2 \) Given revenue function: \( R(x) = 3x \) The profit function P(x) is total revenue minus total cost: \( P(x) = R(x) - C(x) \) \( P(x) = 3x - (100 + 0.015x^2) \) \( P(x) = 3x - 100 - 0.015x^2 \) To find how many items should be produced to maximize profit, take the first derivative of P(x) with respect to x: \( \frac{dP}{dx} = 3 - 0.03x \) Set the first derivative to zero to find critical points: \( 3 - 0.03x = 0 \) \( 3 = 0.03x \) \( x = \frac{3}{0.03} = \frac{3}{\frac{3}{100}} = 3 \times \frac{100}{3} \) \( x = 100 \) Next, use the second derivative to confirm if this value of x gives a maximum. \( \frac{d^2P}{dx^2} = -0.03 \) Since the second derivative is \( -0.03 \), which is less than 0, the profit is maximum when \( x = 100 \). So, 100 items should be produced to maximize profit. Now, calculate the maximum profit at \( x = 100 \): Maximum Profit \( = P(100) = 3(100) - 100 - 0.015(100)^2 \) \( P(100) = 300 - 100 - 0.015(10000) \) \( P(100) = 200 - 150 \) \( P(100) = 50 \) The maximum profit is Rs. 50. Understanding how marginal cost and marginal revenue behave is crucial for finding the optimal production quantity.In simple words: We combine the income and cost formulas into one "profit" formula. Then, we use math to find the perfect number of items to make, so that our profit is the highest it can be. Finally, we calculate what that top profit amount is.

🎯 Exam Tip: Carefully review the given cost and revenue functions to ensure they are correctly integrated into the profit function before performing differentiation.

Question 11. A radio manufacturer finds that he can sell x radios per week at p each, where \( p = 2\left(100-\frac{x}{4}\right) \). His cost of production of x radios per week is \( \left(120 x+\frac{x^2}{2}\right) \). Show that his profit is maximum when the production is 40 radios per week. Find also the maximum profit per week.
Answer: Given price function: \( p = 2\left(100-\frac{x}{4}\right) = 200 - \frac{x}{2} \) Given cost function: \( C(x) = 120x + \frac{x^2}{2} \) First, find the total revenue function R(x): \( R(x) = px = \left(200 - \frac{x}{2}\right)x = 200x - \frac{x^2}{2} \) Next, find the profit function P(x), which is total revenue minus total cost: \( P(x) = R(x) - C(x) \) \( P(x) = \left(200x - \frac{x^2}{2}\right) - \left(120x + \frac{x^2}{2}\right) \) \( P(x) = 200x - \frac{x^2}{2} - 120x - \frac{x^2}{2} \) Combine like terms: \( P(x) = (200x - 120x) + \left(-\frac{x^2}{2} - \frac{x^2}{2}\right) \) \( P(x) = 80x - x^2 \) To find the production level that maximizes profit, take the first derivative of P(x) with respect to x: \( \frac{dP}{dx} = 80 - 2x \) Set the first derivative to zero to find critical points: \( 80 - 2x = 0 \) \( 2x = 80 \) \( x = 40 \) Next, use the second derivative to confirm if this value of x gives a maximum. \( \frac{d^2P}{dx^2} = -2 \) Since the second derivative is \( -2 \), which is less than 0, the profit is maximum when \( x = 40 \). This confirms that profit is maximum when the production is 40 radios per week. Now, find the maximum profit per week by substituting \( x = 40 \) into the profit function: Maximum Profit \( = P(40) = 80(40) - (40)^2 \) \( P(40) = 3200 - 1600 \) \( P(40) = 1600 \) The maximum profit per week is Rs. 1600. When marginal cost equals marginal revenue, profit is maximized, which is a fundamental economic principle.In simple words: We are given formulas for how much money is earned and how much it costs to make radios. We combine these into a "profit" formula. Then, we use special math tools to find the exact number of radios to produce that will give us the biggest profit. Finally, we calculate what that maximum profit amount is.

🎯 Exam Tip: Remember to carefully combine terms in the profit function before differentiating, as errors in this step can lead to incorrect optimization results.

Question 12. (i) Find the profit maximizing output level given \( x = 200 - 10p \) and \( AC = 10 + \frac{x}{25} \), where x represents the units of output, p represents price, and AC represents average cost. (ii) The demand function of an output is \( x = 106 - 2p \), where x is the number of units of output and; the price per unit output. If the total revenue be px, determine the number of units for maximum profit.
Answer: (i) Given demand function: \( x = 200 - 10p \) Rearrange to express p in terms of x: \( 10p = 200 - x \) \( p = \frac{200 - x}{10} = 20 - \frac{x}{10} \) Given average cost (AC): \( AC = 10 + \frac{x}{25} \) Total cost function \( C(x) = AC \times x = \left(10 + \frac{x}{25}\right)x = 10x + \frac{x^2}{25} \) Total revenue function \( R(x) = px = \left(20 - \frac{x}{10}\right)x = 20x - \frac{x^2}{10} \) Profit function \( P(x) = R(x) - C(x) \) \( P(x) = \left(20x - \frac{x^2}{10}\right) - \left(10x + \frac{x^2}{25}\right) \) \( P(x) = 20x - 10x - \frac{x^2}{10} - \frac{x^2}{25} \) \( P(x) = 10x - \left(\frac{5x^2}{50} + \frac{2x^2}{50}\right) \) \( P(x) = 10x - \frac{7x^2}{50} \) To find the profit maximizing output level, take the first derivative of P(x): \( \frac{dP}{dx} = 10 - \frac{14x}{50} = 10 - \frac{7x}{25} \) Set \( \frac{dP}{dx} = 0 \): \( 10 - \frac{7x}{25} = 0 \) \( 10 = \frac{7x}{25} \) \( 7x = 250 \) \( x = \frac{250}{7} \) Next, take the second derivative to confirm maximum: \( \frac{d^2P}{dx^2} = -\frac{7}{25} \) Since \( -\frac{7}{25} < 0 \), this is a maximum. The profit maximizing output level is \( x = \frac{250}{7} \) units. (ii) Given demand function: \( x = 106 - 2p \) Rearrange to express p in terms of x: \( 2p = 106 - x \) \( p = \frac{106 - x}{2} = 53 - \frac{x}{2} \) Let the cost function be \( C(x) = 7x \) (as implied by the solution steps). Total revenue function \( R(x) = px = \left(53 - \frac{x}{2}\right)x = 53x - \frac{x^2}{2} \) Profit function \( P(x) = R(x) - C(x) \) \( P(x) = \left(53x - \frac{x^2}{2}\right) - 7x \) \( P(x) = 46x - \frac{x^2}{2} \) To find the number of units for maximum profit, take the first derivative of P(x): \( \frac{dP}{dx} = 46 - \frac{2x}{2} = 46 - x \) Set \( \frac{dP}{dx} = 0 \): \( 46 - x = 0 \) \( x = 46 \) Next, take the second derivative to confirm maximum: \( \frac{d^2P}{dx^2} = -1 \) Since \( -1 < 0 \), this is a maximum. The number of units for maximum profit is \( x = 46 \) units. The maximum profit is \( P(46) = 46(46) - \frac{(46)^2}{2} = 2116 - \frac{2116}{2} = 2116 - 1058 = 1058 \) Rs. Average cost helps determine the total cost, which is essential for calculating profit.In simple words: This question has two parts. For each part, we first find the formulas for how much money we get (revenue) and how much we spend (cost). We subtract cost from revenue to get the profit. Then, using math, we find the exact number of items that will make the most profit. For the second part, we also calculate what that highest profit amount will be.

🎯 Exam Tip: When a question has multiple parts, clearly label your solutions for each part (i) and (ii) to ensure all aspects are addressed, and verify any implied information from the solution.

Question 13. The cost function C(x) for producing x units of a commodity is given by \( C(x) = \frac{1}{3}x^3 - 5x^2 + 75x + 10 \). At what level of output the marginal cost \( \left(\text{i.e., } \frac{dC}{dx}\right) \) attains its minimum? What is the marginal cost at this level of production?
Answer: Given cost function: \( C(x) = \frac{1}{3}x^3 - 5x^2 + 75x + 10 \) First, find the marginal cost function (MC), which is the first derivative of the total cost function: \( MC(x) = \frac{dC}{dx} = \frac{1}{3}(3x^2) - 5(2x) + 75 \) \( MC(x) = x^2 - 10x + 75 \) To find the minimum marginal cost, we need to take the derivative of the MC function and set it to zero. First derivative of MC: \( \frac{d}{dx}(MC) = 2x - 10 \) Set the first derivative of MC to zero: \( 2x - 10 = 0 \) \( 2x = 10 \) \( x = 5 \) Next, use the second derivative of MC to confirm if this value of x gives a minimum. Second derivative of MC: \( \frac{d^2}{dx^2}(MC) = 2 \) Since the second derivative is 2, which is greater than 0, the marginal cost is minimum when \( x = 5 \). So, the marginal cost attains its minimum at an output level of 5 units. Now, calculate the marginal cost at this level of production \( (x = 5) \): Minimum Marginal Cost \( = MC(5) = (5)^2 - 10(5) + 75 \) \( = 25 - 50 + 75 \) \( = 50 \) The marginal cost at this level of production is Rs. 50. Marginal cost usually decreases initially due to economies of scale, then increases as production capacity is stretched.In simple words: We start with a formula for total cost. We then find a formula for "marginal cost", which is how much extra it costs to make one more item. To find the lowest point of this extra cost, we use another math step. Once we find the number of items where this extra cost is lowest, we calculate that lowest extra cost.

🎯 Exam Tip: Remember that to minimize a function (like Marginal Cost), you must take its derivative, set it to zero to find critical points, and then use the second derivative test to confirm it's a minimum.

Question 14. The total cost function of producing and marketing x units of a commodity is given by \( C = 16 - 12x + 2x^2 \). Find the level of output at which it is minimum.
Answer: Given total cost function: \( C(x) = 16 - 12x + 2x^2 \) To find the level of output at which the total cost is minimum, we take the first derivative of C(x) with respect to x: \( \frac{dC}{dx} = -12 + 2(2x) = -12 + 4x \) Set the first derivative to zero to find critical points: \( -12 + 4x = 0 \) \( 4x = 12 \) \( x = 3 \) Next, use the second derivative to confirm if this value of x gives a minimum. \( \frac{d^2C}{dx^2} = 4 \) Since the second derivative is 4, which is greater than 0, the total cost is minimum when \( x = 3 \). The level of output at which total cost is minimum is 3 units. Businesses aim to produce at the lowest possible total cost to maximize efficiency and potential profit.In simple words: We have a formula for the total cost of making items. To find the lowest cost, we use a special math step to find the exact number of items that should be made. We also check our answer to make sure it's truly the lowest cost.

🎯 Exam Tip: For quadratic cost functions, the minimum (or maximum) often occurs at the vertex, which is efficiently found by setting the first derivative to zero and checking the second derivative.

Question 15. The total cost function of a product is given by \( C(x) = x^3 - 315x^2 + 27000x + 20000 \) where x is the number of units produced. Determine the number of units that should be produced to minimize the total cost.
Answer: Given total cost function: \( C(x) = x^3 - 315x^2 + 27000x + 20000 \) To determine the number of units that should be produced to minimize the total cost, we take the first derivative of C(x) with respect to x: \( \frac{dC}{dx} = 3x^2 - 315(2x) + 27000 \) \( \frac{dC}{dx} = 3x^2 - 630x + 27000 \) Set the first derivative to zero to find critical points: \( 3x^2 - 630x + 27000 = 0 \) Divide the entire equation by 3: \( x^2 - 210x + 9000 = 0 \) Now, factorize this quadratic equation or use the quadratic formula. We look for two numbers that multiply to 9000 and add up to 210. These numbers are 60 and 150. \( (x - 60)(x - 150) = 0 \) So, we have two possible values for x: \( x = 60 \) or \( x = 150 \). Next, use the second derivative to confirm which of these values gives a minimum. \( \frac{d^2C}{dx^2} = 3(2x) - 630 = 6x - 630 \) At \( x = 60 \): \( \frac{d^2C}{dx^2}\Big|_{x=60} = 6(60) - 630 = 360 - 630 = -270 \) Since \( -270 \) is less than 0, \( x = 60 \) corresponds to a local maximum, not a minimum. At \( x = 150 \): \( \frac{d^2C}{dx^2}\Big|_{x=150} = 6(150) - 630 = 900 - 630 = 270 \) Since \( 270 \) is greater than 0, \( x = 150 \) corresponds to a local minimum. Therefore, the number of units that should be produced to minimize the total cost is 150 units. For cost functions that are cubic, there can be both a local maximum and a local minimum, so testing both critical points is important.In simple words: We have a complex formula for the total cost of making items. To find the lowest cost, we use math steps to find the possible numbers of items that could give a low cost. We check each of these numbers to find the one that truly makes the cost the lowest.

🎯 Exam Tip: Always perform the second derivative test for all critical points when you have more than one potential minimum or maximum to correctly identify the true minimum.

Question 16. The manufacturing cost of an item consists of Rs. 1000 as overheads, material cost Rs. 2 per unit for x units and the labour cost \( \frac{x^2}{90} \) for x units produced. Find how many units of the item should be produced so that the average cost is minimum.
Answer: Given fixed overhead cost (TFC) = Rs. 1000. Material cost for x units = \( 2x \) Rs. Labour cost for x units = \( \frac{x^2}{90} \) Rs. The total variable cost (TVC) is the sum of material and labour costs: \( TVC(x) = 2x + \frac{x^2}{90} \) The total cost function C(x) is the sum of total fixed cost and total variable cost: \( C(x) = TFC + TVC(x) = 1000 + 2x + \frac{x^2}{90} \) The average cost function (AC) is total cost divided by the number of units x: \( AC(x) = \frac{C(x)}{x} = \frac{1000 + 2x + \frac{x^2}{90}}{x} \) \( AC(x) = \frac{1000}{x} + 2 + \frac{x}{90} \) To find the minimum average cost, we take the first derivative of AC(x) with respect to x: \( \frac{d}{dx}(AC) = -\frac{1000}{x^2} + 0 + \frac{1}{90} \) \( \frac{d}{dx}(AC) = -\frac{1000}{x^2} + \frac{1}{90} \) Set the first derivative to zero to find critical points: \( -\frac{1000}{x^2} + \frac{1}{90} = 0 \) \( \frac{1}{90} = \frac{1000}{x^2} \) \( x^2 = 1000 \times 90 \) \( x^2 = 90000 \) \( x = \sqrt{90000} \) \( x = 300 \) (Since the number of units x must be positive). Next, use the second derivative to confirm if this value of x gives a minimum. \( \frac{d^2}{dx^2}(AC) = \frac{d}{dx}\left(-1000x^{-2} + \frac{1}{90}\right) \) \( \frac{d^2}{dx^2}(AC) = -1000(-2)x^{-3} = \frac{2000}{x^3} \) Evaluate the second derivative at \( x = 300 \): \( \frac{d^2}{dx^2}(AC)\Big|_{x=300} = \frac{2000}{(300)^3} \) Since \( \frac{2000}{(300)^3} \) is greater than 0, the average cost is minimum when \( x = 300 \). Thus, 300 units of the item should be produced per day to minimize the average cost. Companies often aim to produce at the minimum average cost to achieve operational efficiency.In simple words: We add up all the costs (fixed, material, labor) to get the total cost. Then, we find the "average cost" per item by dividing the total cost by the number of items. Using math, we find the exact number of items to make so that this average cost is the lowest it can be.

🎯 Exam Tip: Remember to calculate the total cost function first, then derive the average cost function by dividing by 'x' before applying differentiation for minimization.

Question 17. The cost function of a firm is \( C = 5x^2 + 28x + 5 \), where C is the cost and x is level of output. A tax at the rate of Rs. 2 per unit of output is imposed and the producer adds it to his cost. Find the minimum value of the average cost.
Answer: Given original cost function: \( C_{orig}(x) = 5x^2 + 28x + 5 \) A tax of Rs. 2 per unit of output is imposed. Total tax for x units of output = \( 2x \) Rs. The producer adds this tax to his cost, so the new total cost function C(x) is: \( C(x) = C_{orig}(x) + 2x = (5x^2 + 28x + 5) + 2x \) \( C(x) = 5x^2 + 30x + 5 \) Now, find the average cost function (AC), which is the total cost divided by the number of units x: \( AC(x) = \frac{C(x)}{x} = \frac{5x^2 + 30x + 5}{x} \) \( AC(x) = 5x + 30 + \frac{5}{x} \) To find the minimum value of the average cost, take the first derivative of AC(x) with respect to x: \( \frac{d}{dx}(AC) = 5 + 0 - \frac{5}{x^2} \) \( \frac{d}{dx}(AC) = 5 - \frac{5}{x^2} \) Set the first derivative to zero to find critical points: \( 5 - \frac{5}{x^2} = 0 \) \( 5 = \frac{5}{x^2} \) \( x^2 = 1 \) \( x = 1 \) (Since the level of output x must be positive). Next, use the second derivative to confirm if this value of x gives a minimum. \( \frac{d^2}{dx^2}(AC) = \frac{d}{dx}(5 - 5x^{-2}) \) \( \frac{d^2}{dx^2}(AC) = 0 - 5(-2)x^{-3} = \frac{10}{x^3} \) Evaluate the second derivative at \( x = 1 \): \( \frac{d^2}{dx^2}(AC)\Big|_{x=1} = \frac{10}{(1)^3} = 10 \) Since 10 is greater than 0, the average cost is minimum when \( x = 1 \). Now, calculate the minimum value of the average cost at \( x = 1 \): Minimum AC \( = AC(1) = 5(1) + 30 + \frac{5}{1} \) \( = 5 + 30 + 5 \) \( = 40 \) The minimum value of the average cost is Rs. 40. Taxes can shift a company's cost curve, potentially changing the optimal production level for minimum average cost.In simple words: We start with the cost formula and then add the new tax cost to it. We then find the "average cost" per item. Using math, we find the exact number of items to make so that this average cost is the lowest. Finally, we calculate what that lowest average cost is.

🎯 Exam Tip: Always update the total cost function to include any new costs, such as taxes, before calculating the average cost and performing differentiation for minimization.

Question 18. The manufacturing cost of an article involves a fixed overhead of Rs. 100 per day, Rs. 0.50 for material per unit, and \( \frac{x^2}{100} \) per day for labour and machinery to produce x articles. How many articles should be produced per day to minimize the average cost per article?
Answer: Fixed overhead cost (TFC) = Rs. 100. Material cost for x articles = \( 0.50x \) Rs. Labour and machinery cost for x articles = \( \frac{x^2}{100} \) Rs. The total variable cost (TVC) is the sum of material and labour/machinery costs: \( TVC(x) = 0.50x + \frac{x^2}{100} \) The total cost function C(x) is the sum of total fixed cost and total variable cost: \( C(x) = TFC + TVC(x) = 100 + 0.50x + \frac{x^2}{100} \) The average cost function (AC) is total cost divided by the number of articles x: \( AC(x) = \frac{C(x)}{x} = \frac{100 + 0.50x + \frac{x^2}{100}}{x} \) \( AC(x) = \frac{100}{x} + 0.50 + \frac{x}{100} \) To find how many articles should be produced to minimize the average cost, take the first derivative of AC(x) with respect to x: \( \frac{d}{dx}(AC) = -\frac{100}{x^2} + 0 + \frac{1}{100} \) \( \frac{d}{dx}(AC) = -\frac{100}{x^2} + \frac{1}{100} \) Set the first derivative to zero to find critical points: \( -\frac{100}{x^2} + \frac{1}{100} = 0 \) \( \frac{1}{100} = \frac{100}{x^2} \) \( x^2 = 100 \times 100 \) \( x^2 = 10000 \) \( x = \sqrt{10000} \) \( x = 100 \) (Since the number of articles x must be positive). Next, use the second derivative to confirm if this value of x gives a minimum. \( \frac{d^2}{dx^2}(AC) = \frac{d}{dx}\left(-100x^{-2} + \frac{1}{100}\right) \) \( \frac{d^2}{dx^2}(AC) = -100(-2)x^{-3} = \frac{200}{x^3} \) Evaluate the second derivative at \( x = 100 \): \( \frac{d^2}{dx^2}(AC)\Big|_{x=100} = \frac{200}{(100)^3} \) Since \( \frac{200}{(100)^3} \) is greater than 0, the average cost is minimum when \( x = 100 \). Thus, 100 articles should be produced per day to minimize the average cost per article. Minimizing average cost helps a company produce goods most efficiently, often leading to competitive pricing.In simple words: We calculate the total cost for making articles, including fixed costs and costs that change with production. Then, we find the average cost per article. Using math, we determine the exact number of articles to produce daily to make this average cost as low as possible.

🎯 Exam Tip: Pay close attention to how variable costs are expressed in the problem statement (e.g., per unit, as a squared term) to correctly formulate the total cost function.

Question 19. (i) A firm produces x units of output per week at a total cost of \( \frac{x^3}{3} - x^2 + 5x + 3 \), find the output levels at which the marginal cost and average variable cost attain their respective minima. (ii) The cost function of a firm is given by \( C = \frac{1}{3}x^3 - 5x^2 + 30x + 10 \), where C is the total cost for x items. Determine x at which the marginal cost is minimum.
Answer: (i) Given total cost function: \( C(x) = \frac{x^3}{3} - x^2 + 5x + 3 \) **For Marginal Cost (MC) minimum:** First, find the marginal cost function: \( MC(x) = \frac{dC}{dx} = \frac{1}{3}(3x^2) - 2x + 5 = x^2 - 2x + 5 \) To find the minimum MC, take the derivative of MC: \( \frac{d}{dx}(MC) = 2x - 2 \) Set this to zero: \( 2x - 2 = 0 \Rightarrow 2x = 2 \Rightarrow x = 1 \) Second derivative of MC: \( \frac{d^2}{dx^2}(MC) = 2 \) Since \( 2 > 0 \), MC is minimum at \( x = 1 \) unit. **For Average Variable Cost (AVC) minimum:** First, find the total variable cost (TVC). Fixed cost (FC) is \( C(0) = 3 \). \( TVC(x) = C(x) - FC = \left(\frac{x^3}{3} - x^2 + 5x + 3\right) - 3 = \frac{x^3}{3} - x^2 + 5x \) Now, find the average variable cost function: \( AVC(x) = \frac{TVC(x)}{x} = \frac{\frac{x^3}{3} - x^2 + 5x}{x} = \frac{x^2}{3} - x + 5 \) To find the minimum AVC, take the derivative of AVC: \( \frac{d}{dx}(AVC) = \frac{2x}{3} - 1 \) Set this to zero: \( \frac{2x}{3} - 1 = 0 \Rightarrow \frac{2x}{3} = 1 \Rightarrow x = \frac{3}{2} \) Second derivative of AVC: \( \frac{d^2}{dx^2}(AVC) = \frac{2}{3} \) Since \( \frac{2}{3} > 0 \), AVC is minimum at \( x = \frac{3}{2} \) units. (ii) Given cost function: \( C(x) = \frac{1}{3}x^3 - 5x^2 + 30x + 10 \) **For Marginal Cost (MC) minimum:** First, find the marginal cost function: \( MC(x) = \frac{dC}{dx} = \frac{1}{3}(3x^2) - 5(2x) + 30 = x^2 - 10x + 30 \) To find the minimum MC, take the derivative of MC: \( \frac{d}{dx}(MC) = 2x - 10 \) Set this to zero: \( 2x - 10 = 0 \Rightarrow 2x = 10 \Rightarrow x = 5 \) Second derivative of MC: \( \frac{d^2}{dx^2}(MC) = 2 \) Since \( 2 > 0 \), MC is minimum at \( x = 5 \) units. The marginal cost curve intersects the average variable cost curve at the AVC's minimum point.In simple words: In two parts, we are given cost formulas. For part one, we find the formula for "extra cost" (marginal cost) and "average changing cost" (average variable cost), then use math to find the number of items where each of these costs is lowest. For part two, we do the same for the "extra cost" to find its lowest point.

🎯 Exam Tip: Remember that fixed cost does not affect marginal cost, but it is necessary to calculate total variable cost to then find average variable cost.

Question 20. Given a quadratic cost function \( C(x) = ax^2 + bx + c \), minimize the average cost and show that the average cost is equal to marginal cost at that value.
Answer: Given quadratic cost function: \( C(x) = ax^2 + bx + c \) First, find the average cost function (AC): \( AC(x) = \frac{C(x)}{x} = \frac{ax^2 + bx + c}{x} = ax + b + \frac{c}{x} \) To minimize AC, take the first derivative of AC(x) with respect to x: \( \frac{d}{dx}(AC) = a - \frac{c}{x^2} \) Set the first derivative to zero to find the critical point: \( a - \frac{c}{x^2} = 0 \) \( a = \frac{c}{x^2} \) \( x^2 = \frac{c}{a} \) \( x = \sqrt{\frac{c}{a}} \) (Assuming \( a, c > 0 \) for real positive output and a minimum). Next, use the second derivative to confirm if this value of x gives a minimum. \( \frac{d^2}{dx^2}(AC) = \frac{d}{dx}(ax - cx^{-2}) = a - c(-2)x^{-3} = \frac{2c}{x^3} \) Evaluate the second derivative at \( x = \sqrt{\frac{c}{a}} \): \( \frac{d^2}{dx^2}(AC)\Big|_{x=\sqrt{\frac{c}{a}}} = \frac{2c}{(\sqrt{c/a})^3} = \frac{2c}{(c/a)\sqrt{c/a}} = \frac{2a}{\sqrt{c/a}} = 2a\sqrt{\frac{a}{c}} \) Since \( a, c > 0 \), \( 2a\sqrt{\frac{a}{c}} > 0 \), so the average cost is minimum at \( x = \sqrt{\frac{c}{a}} \). Now, calculate the minimum average cost at this output level: Minimum AC \( = AC\left(\sqrt{\frac{c}{a}}\right) = a\sqrt{\frac{c}{a}} + b + \frac{c}{\sqrt{c/a}} \) \( = a\frac{\sqrt{c}}{\sqrt{a}} + b + \frac{c\sqrt{a}}{\sqrt{c}} = \sqrt{a^2 \frac{c}{a}} + b + \sqrt{c^2 \frac{a}{c}} \) \( = \sqrt{ac} + b + \sqrt{ac} = b + 2\sqrt{ac} \) Next, we show that marginal cost (MC) is equal to average cost (AC) at this value of x. Find the marginal cost function: \( MC(x) = \frac{dC}{dx} = 2ax + b \) Evaluate MC at \( x = \sqrt{\frac{c}{a}} \): \( MC\left(\sqrt{\frac{c}{a}}\right) = 2a\sqrt{\frac{c}{a}} + b = 2\sqrt{a^2 \frac{c}{a}} + b = 2\sqrt{ac} + b \) Comparing the minimum AC and the MC at \( x = \sqrt{\frac{c}{a}} \): Minimum AC \( = b + 2\sqrt{ac} \) MC at minimum AC \( = b + 2\sqrt{ac} \) Thus, we have shown that Average Cost (AC) is equal to Marginal Cost (MC) at the point of minimum average cost. This fundamental relationship, where MC intersects AC at its lowest point, is a cornerstone of microeconomic theory.In simple words: We start with a general cost formula. We find the "average cost" formula and use math to figure out the number of items that makes this average cost the lowest. Then, we find the "extra cost" (marginal cost) formula. Finally, we show that at the exact number of items where average cost is lowest, the average cost and the extra cost are the same.

🎯 Exam Tip: Remember this key result: Marginal Cost (MC) always equals Average Cost (AC) at the point where Average Cost is at its minimum. This is a crucial concept in economics.