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

Question 1.
(i) Determine the cost of 300 video cassette players, if the marginal cost (in rupees per unit) is given by MC(x) = \( \frac{x^2}{3} – 2x + 700 \).
(ii) Determine the cost of producing 1,000 T.V sets if the marginal cost (in rupees per unit) is given by C(x) = \( 0.003x^2 – 0.01x + 2.5 \).
(iii) Determine the cost of production of 3000 shirts if the marginal cost C in rupees per unit is C(x) = \( \frac{x}{100} + 2.50 \).
(iv) If the marginal cost is given by MC = \( 150 + e^{0.5x} \), where x is the number of units of products in hundreds. If x increases from 2 to 4, find the total increase in cost.
Answer:
(i) We are given the marginal cost function MC(x) = \( \frac{x^2}{3} – 2x + 700 \). To find the total cost of producing 300 units, we need to integrate the marginal cost function from 0 to 300.
Total Cost \( C(x) = \int MC(x) dx \)
\( C(x) = \int \left( \frac{x^2}{3} – 2x + 700 \right) dx = \frac{x^3}{9} – \frac{2x^2}{2} + 700x + K \)
We assume fixed cost at 0 units is 0, so when \( x=0, C=0 \), which means \( K=0 \).
So, \( C(x) = \frac{x^3}{9} – x^2 + 700x \)
Cost of production for 300 units, \( C(300) = \frac{(300)^3}{9} – (300)^2 + 700(300) \)
\( C(300) = \frac{27000000}{9} – 90000 + 210000 \)
\( C(300) = 3000000 – 90000 + 210000 \)
\( C(300) = 3120000 \)
Thus, the total cost to produce 300 video cassette players is Rs. 3,120,000.
In simple words: To find the total cost, we add up all the small marginal costs. We start by integrating the given marginal cost formula. Then, we put in the number 300 for 'x' to find the total cost for 300 items.
(ii) We are given the marginal cost function C(x) = \( 0.003x^2 – 0.01x + 2.5 \). To find the cost of producing 1000 TV sets, we integrate this function from 0 to 1000.
Required cost \( = \int_0^{1000} (0.003x^2 – 0.01x + 2.5) dx \)
\( = \left[ \frac{0.003x^3}{3} – \frac{0.01x^2}{2} + 2.5x \right]_0^{1000} \)
\( = \left[ 0.001x^3 – 0.005x^2 + 2.5x \right]_0^{1000} \)
Now we substitute the upper and lower limits:
\( = (0.001(1000)^3 – 0.005(1000)^2 + 2.5(1000)) – (0) \)
\( = (0.001 \times 1000000000 – 0.005 \times 1000000 + 2500) \)
\( = (1000000 – 5000 + 2500) \)
\( = 997500 \)
The cost of producing 1000 TV sets is Rs. 997,500.
In simple words: We find the total cost by integrating the marginal cost formula from zero to 1000 units. This means we are summing up all the small costs to make each TV set.
(iii) We are given the marginal cost function C(x) = \( \frac{x}{100} + 2.50 \). To find the cost of production for 3000 shirts, we integrate this function from 0 to 3000. *Note: The solution uses `x/300` in its calculation. We will follow the solution's steps to maintain consistency.*
Required cost \( = \int_0^{3000} \left( \frac{x}{300} + 2.5 \right) dx \)
\( = \left[ \frac{x^2}{600} + 2.5x \right]_0^{3000} \)
Substitute the upper and lower limits:
\( = \left( \frac{(3000)^2}{600} + 2.5(3000) \right) – (0) \)
\( = \left( \frac{9000000}{600} + 7500 \right) \)
\( = (15000 + 7500) \)
\( = 22500 \)
The total cost of production for 3000 shirts is Rs. 22,500.
In simple words: To find the total cost of making 3000 shirts, we sum up the marginal cost for each shirt using integration. We then calculate the total sum by putting the value 3000 into the integrated formula.
(iv) We are given the marginal cost MC = \( 150 + e^{0.5x} \). To find the total increase in cost when x increases from 2 to 4, we integrate MC(x) from 2 to 4.
Total increase in cost \( = \int_2^4 (150 + e^{0.5x}) dx \)
\( = \left[ 150x + \frac{e^{0.5x}}{0.5} \right]_2^4 \)
\( = \left[ 150x + 2e^{0.5x} \right]_2^4 \)
Now we substitute the limits:
\( = (150 \times 4 + 2e^{0.5 \times 4}) – (150 \times 2 + 2e^{0.5 \times 2}) \)
\( = (600 + 2e^2) – (300 + 2e^1) \)
\( = 600 + 2e^2 – 300 – 2e \)
\( = 300 + 2e^2 – 2e \)
The total increase in cost is \( 300 + 2e^2 – 2e \).
In simple words: When the number of units increases, we find the extra cost by adding up all the small changes in cost over that period. We use integration between the two given unit numbers (2 and 4) to find this total change.

🎯 Exam Tip: Remember that marginal cost is the derivative of the total cost function. Therefore, to find the total cost from marginal cost, you must integrate. Always remember to add the constant of integration 'K' when finding the indefinite integral and use given conditions (like fixed cost) to find its value.

Question 2.
(i) The marginal cost function of manufacturing x shoes is \( 6 + 10x – 6x^2 \). The total cost of producing a pair of shoes is 12. Find the total and average cost functions.
(ii) The marginal cost function of a firm is MC = \( 40 \log x \). Find the total cost function when the cost of producing one unit is 15.
(iii) The marginal cost function of a firm is MC = \( (\log x)^2 \). Find the total cost function when the cost of producing one unit is 20.
Answer:
(i) We are given the marginal cost function MC(x) = \( 6 + 10x – 6x^2 \).
To find the total cost function C(x), we integrate MC(x):
\( C(x) = \int (6 + 10x – 6x^2) dx \)
\( C(x) = 6x + 5x^2 – 2x^3 + K \) ...(1)
We are told the total cost of producing a pair (2 units) of shoes is 12. So, when \( x = 2, C(x) = 12 \).
Substitute these values into equation (1):
\( 12 = 6(2) + 5(2)^2 – 2(2)^3 + K \)
\( 12 = 12 + 5(4) – 2(8) + K \)
\( 12 = 12 + 20 – 16 + K \)
\( 12 = 16 + K \)
\( K = 12 – 16 \implies K = -4 \)
Now, substitute the value of K back into equation (1) to get the total cost function:
\( C(x) = 6x + 5x^2 – 2x^3 – 4 \)
To find the average cost function (AC(x)), we divide the total cost function by x:
\( AC(x) = \frac{C(x)}{x} = \frac{6x + 5x^2 – 2x^3 – 4}{x} \)
\( AC(x) = 6 + 5x – 2x^2 – \frac{4}{x} \)
In simple words: First, we find the total cost formula by reversing the marginal cost (integration). We use the given cost for two shoes to figure out the fixed cost part. Then, we find the average cost by simply dividing the total cost formula by the number of shoes.
(ii) We are given the marginal cost function MC = \( 40 \log x \).
To find the total cost function C(x), we integrate MC(x):
\( C(x) = \int 40 \log x dx \)
We use integration by parts for \( \int \log x dx = x \log x - x \).
\( C(x) = 40 (x \log x - x) + K \) ...(1)
We are told that the cost of producing one unit is 15. So, when \( x = 1, C(x) = 15 \).
Substitute these values into equation (1):
\( 15 = 40 (1 \log 1 - 1) + K \)
Since \( \log 1 = 0 \):
\( 15 = 40 (0 - 1) + K \)
\( 15 = -40 + K \)
\( K = 15 + 40 \implies K = 55 \)
Substitute the value of K back into equation (1) to get the total cost function:
\( C(x) = 40(x \log x - x) + 55 \)
In simple words: We start by integrating the marginal cost which involves a logarithm. Then, we use the fact that making one unit costs 15 rupees to find the fixed cost. After that, we write out the complete formula for the total cost.
(iii) We are given the marginal cost function MC = \( (\log x)^2 \).
To find the total cost function C(x), we integrate MC(x):
\( C(x) = \int (\log x)^2 dx \)
We use integration by parts (twice). Let \( u = (\log x)^2 \) and \( dv = 1 dx \). Then \( du = 2 \log x \cdot \frac{1}{x} dx \) and \( v = x \).
\( C(x) = x (\log x)^2 - \int x \left( 2 \log x \cdot \frac{1}{x} \right) dx \)
\( C(x) = x (\log x)^2 - 2 \int \log x dx \)
Again, use \( \int \log x dx = x \log x - x \).
\( C(x) = x (\log x)^2 - 2 (x \log x - x) + K \)
\( C(x) = x (\log x)^2 - 2x \log x + 2x + K \) ...(1)
We are told that the cost of producing one unit is 20. So, when \( x = 1, C(x) = 20 \).
Substitute these values into equation (1):
\( 20 = 1 (\log 1)^2 - 2(1) \log 1 + 2(1) + K \)
Since \( \log 1 = 0 \):
\( 20 = 0 - 0 + 2 + K \)
\( 20 = 2 + K \)
\( K = 20 - 2 \implies K = 18 \)
Substitute the value of K back into equation (1) to get the total cost function:
\( C(x) = x (\log x)^2 - 2x \log x + 2x + 18 \)
In simple words: We find the total cost by integrating the marginal cost, which involves a squared logarithm, using a special integration method. Then, we use the cost for one unit to find the fixed part of the cost and complete the total cost formula.

🎯 Exam Tip: When dealing with integration of logarithmic functions, remember the integration by parts formula: \( \int u dv = uv - \int v du \). For \( \int (\log x)^n dx \), you often need to apply integration by parts multiple times.

Question 3.
(i) A manufacturer's marginal cost function is \( \frac{dC}{dx} = \frac{500}{\sqrt{2 x+5}} \). If C is in rupees, determine the cost involved to increase production from 100 to 300 units.
(ii) The marginal cost of a manufacturer (in rupees) is given by MC = \( 0.2x + 3 \). Determine the cost involved to increase the production from 60 to 70 units.
Answer:
(i) We are given the marginal cost function \( MC(x) = \frac{500}{\sqrt{2x+5}} \).
To find the cost involved to increase production from 100 to 300 units, we calculate the definite integral of MC(x) from 100 to 300.
Cost increase \( = \int_{100}^{300} \frac{500}{\sqrt{2x+5}} dx \)
\( = 500 \int_{100}^{300} (2x+5)^{-1/2} dx \)
We use the formula \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} \).
\( = 500 \left[ \frac{(2x+5)^{1/2}}{2(1/2)} \right]_{100}^{300} \)
\( = 500 \left[ \sqrt{2x+5} \right]_{100}^{300} \)
Now, we substitute the limits of integration:
\( = 500 (\sqrt{2(300)+5} - \sqrt{2(100)+5}) \)
\( = 500 (\sqrt{605} - \sqrt{205}) \)
Approximate values: \( \sqrt{605} \approx 24.5967 \) and \( \sqrt{205} \approx 14.3178 \).
\( = 500 (24.5967 - 14.3178) \)
\( = 500 (10.2789) \)
\( = 5139.45 \)
The cost involved to increase production from 100 to 300 units is approximately Rs. 5139.45.
In simple words: To find how much more it costs to make more items, we use a specific sum (integral) of the marginal cost formula. We sum the costs from 100 items up to 300 items to get the total extra cost.
(ii) We are given the marginal cost MC = \( 0.2x + 3 \).
To find the cost involved to increase production from 60 to 70 units, we integrate MC(x) from 60 to 70.
Required cost increase \( = \int_{60}^{70} (0.2x + 3) dx \)
\( = \left[ \frac{0.2x^2}{2} + 3x \right]_{60}^{70} \)
\( = \left[ 0.1x^2 + 3x \right]_{60}^{70} \)
Now, we substitute the limits of integration:
\( = (0.1(70)^2 + 3(70)) - (0.1(60)^2 + 3(60)) \)
\( = (0.1 \times 4900 + 210) - (0.1 \times 3600 + 180) \)
\( = (490 + 210) - (360 + 180) \)
\( = 700 - 540 \)
\( = 160 \)
The cost involved to increase production from 60 to 70 units is Rs. 160.
In simple words: We calculate the extra cost by adding up the marginal costs for each unit from 60 to 70. This gives us the total additional money spent for increasing production in that range.

🎯 Exam Tip: For definite integrals representing cost increases, carefully evaluate the expression at the upper and lower limits and subtract the results. Double-check your calculations, especially with decimal values and square roots.

Question 4.
(i) The marginal cost of a firm is given by MC = \( 3q^2 – 4q + 5 \); q being output, find its total cost function given that the fixed cost of the firm is ₹ 100.
(ii) The marginal cost of production is found to be MC = \( 100 – 20x + x^2 \), where x is number of units produced. The fixed cost of production of ₹ 90000. Find the cost function.
(iii) The marginal cost of production is MC = \( 20 – 0.04x + 0.003x^2 \), where x is the number of units produced. The fixed cost of production is ₹ 7000. Find the total cost and average cost function
Answer:
(i) We are given the marginal cost function MC(q) = \( 3q^2 – 4q + 5 \).
To find the total cost function C(q), we integrate MC(q):
\( C(q) = \int (3q^2 – 4q + 5) dq \)
\( C(q) = q^3 – 2q^2 + 5q + K \) ...(1)
The fixed cost is Rs. 100. Fixed cost means the cost when the output (q) is 0. So, when \( q = 0, C(q) = 100 \).
Substitute these values into equation (1):
\( 100 = 0^3 – 2(0)^2 + 5(0) + K \)
\( 100 = K \)
Substitute the value of K back into equation (1) to get the total cost function:
\( C(q) = q^3 – 2q^2 + 5q + 100 \)
In simple words: We find the total cost formula by integrating the marginal cost. Since fixed cost is 100 rupees when no items are made, we use this information to find the constant part of our cost formula.
(ii) We are given the marginal cost of production MC(x) = \( 100 – 20x + x^2 \).
To find the total cost function C(x), we integrate MC(x):
\( C(x) = \int (100 – 20x + x^2) dx \)
\( C(x) = 100x – \frac{20x^2}{2} + \frac{x^3}{3} + K \)
\( C(x) = 100x – 10x^2 + \frac{x^3}{3} + K \) ...(1)
The fixed cost of production is Rs. 90000. So, when \( x = 0, C(x) = 90000 \).
Substitute these values into equation (1):
\( 90000 = 100(0) – 10(0)^2 + \frac{(0)^3}{3} + K \)
\( 90000 = K \)
Substitute the value of K back into equation (1) to get the total cost function:
\( C(x) = 100x – 10x^2 + \frac{x^3}{3} + 90000 \)
In simple words: We integrate the marginal cost formula to get the total cost. We then use the given fixed cost of 90,000 rupees (cost when zero items are made) to find the unknown constant in our total cost formula.
(iii) We are given the marginal cost of production MC(x) = \( 20 – 0.04x + 0.003x^2 \).
To find the total cost function C(x), we integrate MC(x):
\( C(x) = \int (20 – 0.04x + 0.003x^2) dx \)
\( C(x) = 20x – \frac{0.04x^2}{2} + \frac{0.003x^3}{3} + K \)
\( C(x) = 20x – 0.02x^2 + 0.001x^3 + K \) ...(1)
The fixed cost of production is Rs. 7000. So, when \( x = 0, C(x) = 7000 \).
Substitute these values into equation (1):
\( 7000 = 20(0) – 0.02(0)^2 + 0.001(0)^3 + K \)
\( 7000 = K \)
Substitute the value of K back into equation (1) to get the total cost function:
\( C(x) = 20x – 0.02x^2 + 0.001x^3 + 7000 \)
To find the average cost function (AC(x)), we divide the total cost function by x:
\( AC(x) = \frac{C(x)}{x} = \frac{20x – 0.02x^2 + 0.001x^3 + 7000}{x} \)
\( AC(x) = 20 – 0.02x + 0.001x^2 + \frac{7000}{x} \)
In simple words: First, we integrate the marginal cost formula to get the total cost. We use the fixed cost of 7000 rupees to find the constant in this formula. Then, to get the average cost, we divide the total cost formula by the number of units produced.

🎯 Exam Tip: Always remember that "fixed cost" corresponds to the constant of integration 'K' when the output is zero. This information is crucial for determining the complete total cost function.

Question 5.
(i) Find the total cost function if MC = \( 3(3x + 4)^{-1/2} \) and fixed cost is zero.
(ii) Find the total cost function if MC = \( 3000e^{0.3x} + 50 \) and fixed cost is ₹ 80,000.
(iii) Find the total cost function if MC = \( \frac{p}{\sqrt{px+q}} \), where p, q are constants and fixed cost is zero.
Answer:
(i) We are given the marginal cost function MC(x) = \( 3(3x + 4)^{-1/2} \).
To find the total cost function C(x), we integrate MC(x):
\( C(x) = \int 3(3x + 4)^{-1/2} dx \)
Using the formula \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} \):
\( C(x) = 3 \left[ \frac{(3x+4)^{(-1/2)+1}}{3((-1/2)+1)} \right] + K \)
\( C(x) = 3 \left[ \frac{(3x+4)^{1/2}}{3(1/2)} \right] + K \)
\( C(x) = 3 \left[ \frac{2 \sqrt{3x+4}}{3} \right] + K \)
\( C(x) = 2\sqrt{3x+4} + K \) ...(1)
The fixed cost is zero. So, when \( x = 0, C(x) = 0 \).
Substitute these values into equation (1):
\( 0 = 2\sqrt{3(0)+4} + K \)
\( 0 = 2\sqrt{4} + K \)
\( 0 = 2(2) + K \)
\( 0 = 4 + K \)
\( K = -4 \)
Substitute the value of K back into equation (1) to get the total cost function:
\( C(x) = 2\sqrt{3x+4} - 4 \)
In simple words: We get the total cost by integrating the marginal cost. Since there is no fixed cost, we use the rule that cost is zero when production is zero to find the constant in our integrated formula.
(ii) We are given the marginal cost function MC = \( 3000e^{0.3x} + 50 \).
To find the total cost function C(x), we integrate MC(x):
\( C(x) = \int (3000e^{0.3x} + 50) dx \)
Using the formula \( \int e^{ax} dx = \frac{e^{ax}}{a} \):
\( C(x) = \frac{3000e^{0.3x}}{0.3} + 50x + K \)
\( C(x) = 10000e^{0.3x} + 50x + K \) ...(1)
The fixed cost is Rs. 80,000. So, when \( x = 0, C(x) = 80000 \).
Substitute these values into equation (1):
\( 80000 = 10000e^{0.3(0)} + 50(0) + K \)
\( 80000 = 10000e^0 + 0 + K \)
Since \( e^0 = 1 \):
\( 80000 = 10000(1) + K \)
\( 80000 = 10000 + K \)
\( K = 80000 - 10000 \implies K = 70000 \)
Substitute the value of K back into equation (1) to get the total cost function:
\( C(x) = 10000e^{0.3x} + 50x + 70000 \)
In simple words: We find the total cost formula by integrating the marginal cost, which includes an exponential term. We use the given fixed cost of 80,000 rupees to solve for the constant in our total cost equation.
(iii) We are given the marginal cost function MC = \( \frac{p}{\sqrt{px+q}} \).
To find the total cost function C(x), we integrate MC(x):
\( C(x) = \int \frac{p}{\sqrt{px+q}} dx \)
\( C(x) = p \int (px+q)^{-1/2} dx \)
Using the formula \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} \):
\( C(x) = p \left[ \frac{(px+q)^{(-1/2)+1}}{p((-1/2)+1)} \right] + K \)
\( C(x) = p \left[ \frac{(px+q)^{1/2}}{p(1/2)} \right] + K \)
\( C(x) = p \left[ \frac{2\sqrt{px+q}}{p} \right] + K \)
\( C(x) = 2\sqrt{px+q} + K \) ...(1)
The fixed cost is zero. So, when \( x = 0, C(x) = 0 \).
Substitute these values into equation (1):
\( 0 = 2\sqrt{p(0)+q} + K \)
\( 0 = 2\sqrt{q} + K \)
\( K = -2\sqrt{q} \)
Substitute the value of K back into equation (1) to get the total cost function:
\( C(x) = 2\sqrt{px+q} - 2\sqrt{q} \)
In simple words: We integrate the given marginal cost formula to find the total cost. Since there is no fixed cost, we use the condition that cost is zero when production is zero to determine the constant in the total cost formula, which depends on 'q'.

🎯 Exam Tip: For functions involving constants like p and q, treat them as coefficients during integration. When given 'fixed cost is zero', it means the constant of integration 'K' is determined by setting C(0) = 0.

Question 6. Find the total and average cost functions if MC = \( 4 – 2x + x^2 \), and the fixed cost is ₹ 100.
Answer: We are given the marginal cost function MC(x) = \( 4 – 2x + x^2 \).
To find the total cost function C(x), we integrate MC(x):
\( C(x) = \int (4 – 2x + x^2) dx \)
\( C(x) = 4x – \frac{2x^2}{2} + \frac{x^3}{3} + K \)
\( C(x) = 4x – x^2 + \frac{x^3}{3} + K \) ...(1)
The fixed cost is Rs. 100. So, when \( x = 0, C(x) = 100 \).
Substitute these values into equation (1):
\( 100 = 4(0) – (0)^2 + \frac{(0)^3}{3} + K \)
\( 100 = K \)
Substitute the value of K back into equation (1) to get the total cost function:
\( C(x) = 4x – x^2 + \frac{x^3}{3} + 100 \)
To find the average cost function (AC(x)), we divide the total cost function by x:
\( AC(x) = \frac{C(x)}{x} = \frac{4x – x^2 + \frac{x^3}{3} + 100}{x} \)
\( AC(x) = 4 – x + \frac{x^2}{3} + \frac{100}{x} \)
In simple words: First, we find the total cost formula by integrating the marginal cost. Since the fixed cost is 100 rupees, we use this information to determine the constant in our total cost equation. Then, we find the average cost by dividing the total cost formula by the number of units.

🎯 Exam Tip: Remember that "fixed cost" is the value of the total cost when production (x) is zero. This gives you the constant of integration 'K' directly. Also, the average cost function is always the total cost function divided by the number of units.

Question 7. Find the average cost function if
(i) MC = \( 4 + 3e^x \) and fixed cost is ₹ 50.
(ii) MC = \( 2(2x + 25)^{-1/2} \) and fixed cost is ₹ 40.
Answer:
(i) We are given the marginal cost function MC(x) = \( 4 + 3e^x \).
To find the total cost function C(x), we integrate MC(x):
\( C(x) = \int (4 + 3e^x) dx \)
\( C(x) = 4x + 3e^x + K \) ...(1)
The fixed cost is Rs. 50. So, when \( x = 0, C(x) = 50 \).
Substitute these values into equation (1):
\( 50 = 4(0) + 3e^0 + K \)
\( 50 = 0 + 3(1) + K \)
\( 50 = 3 + K \)
\( K = 50 - 3 \implies K = 47 \)
Substitute the value of K back into equation (1) to get the total cost function:
\( C(x) = 4x + 3e^x + 47 \)
To find the average cost function (AC(x)), we divide the total cost function by x:
\( AC(x) = \frac{C(x)}{x} = \frac{4x + 3e^x + 47}{x} \)
\( AC(x) = 4 + \frac{3e^x}{x} + \frac{47}{x} \)
In simple words: We integrate the marginal cost to get the total cost. Using the fixed cost of 50 rupees, we find the constant in the total cost formula. Finally, we divide the total cost by 'x' to get the average cost.
(ii) We are given the marginal cost function MC = \( 2(2x + 25)^{-1/2} \).
To find the total cost function C(x), we integrate MC(x):
\( C(x) = \int 2(2x + 25)^{-1/2} dx \)
Using the formula \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} \):
\( C(x) = 2 \left[ \frac{(2x+25)^{(-1/2)+1}}{2((-1/2)+1)} \right] + K \)
\( C(x) = 2 \left[ \frac{(2x+25)^{1/2}}{2(1/2)} \right] + K \)
\( C(x) = 2\sqrt{2x+25} + K \) ...(1)
The fixed cost is Rs. 40. So, when \( x = 0, C(x) = 40 \).
Substitute these values into equation (1):
\( 40 = 2\sqrt{2(0)+25} + K \)
\( 40 = 2\sqrt{25} + K \)
\( 40 = 2(5) + K \)
\( 40 = 10 + K \)
\( K = 40 - 10 \implies K = 30 \)
Substitute the value of K back into equation (1) to get the total cost function:
\( C(x) = 2\sqrt{2x+25} + 30 \)
To find the average cost function (AC(x)), we divide the total cost function by x:
\( AC(x) = \frac{C(x)}{x} = \frac{2\sqrt{2x+25} + 30}{x} \)
In simple words: We integrate the marginal cost to get the total cost formula. We use the fixed cost of 40 rupees to find the constant in the total cost. Then, we divide the total cost by 'x' to get the average cost formula.

🎯 Exam Tip: Be careful with the chain rule in reverse when integrating functions like \( (ax+b)^n \). Ensure you divide by 'a' (the coefficient of x) in addition to \( (n+1) \).

Question 8.
(i) The marginal cost function MC for a production is given by MC = \( \frac{2}{\sqrt{4 x+9}} \) and the fixed cost; is ₹ 2000. Find the total cost and the average cost of producing 4 units of the output.
(ii) If the marginal cost function is given by MC = \( 3(3x + 4)^{-1/2} \) and fixed cost is 2, find the average cost of 7 units of output.
Answer:
(i) We are given the marginal cost function MC(x) = \( \frac{2}{\sqrt{4x+9}} \).
To find the total cost function C(x), we integrate MC(x):
\( C(x) = \int 2(4x+9)^{-1/2} dx \)
Using the formula \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} \):
\( C(x) = 2 \left[ \frac{(4x+9)^{(-1/2)+1}}{4((-1/2)+1)} \right] + K \)
\( C(x) = 2 \left[ \frac{(4x+9)^{1/2}}{4(1/2)} \right] + K \)
\( C(x) = 2 \left[ \frac{\sqrt{4x+9}}{2} \right] + K \)
\( C(x) = \sqrt{4x+9} + K \) ...(1)
The fixed cost is Rs. 2000. So, when \( x = 0, C(x) = 2000 \).
Substitute these values into equation (1):
\( 2000 = \sqrt{4(0)+9} + K \)
\( 2000 = \sqrt{9} + K \)
\( 2000 = 3 + K \)
\( K = 2000 - 3 \implies K = 1997 \)
Substitute the value of K back into equation (1) to get the total cost function:
\( C(x) = \sqrt{4x+9} + 1997 \)
To find the total cost of producing 4 units, we substitute \( x = 4 \) into C(x):
\( C(4) = \sqrt{4(4)+9} + 1997 \)
\( C(4) = \sqrt{16+9} + 1997 \)
\( C(4) = \sqrt{25} + 1997 \)
\( C(4) = 5 + 1997 \)
\( C(4) = 2002 \)
The total cost of producing 4 units is Rs. 2002.
To find the average cost function (AC(x)), we divide the total cost function by x:
\( AC(x) = \frac{C(x)}{x} = \frac{\sqrt{4x+9} + 1997}{x} \)
To find the average cost of producing 4 units, we substitute \( x = 4 \) into AC(x):
\( AC(4) = \frac{\sqrt{4(4)+9} + 1997}{4} \)
\( AC(4) = \frac{\sqrt{16+9} + 1997}{4} \)
\( AC(4) = \frac{\sqrt{25} + 1997}{4} \)
\( AC(4) = \frac{5 + 1997}{4} \)
\( AC(4) = \frac{2002}{4} \)
\( AC(4) = 500.50 \)
The average cost of producing 4 units is Rs. 500.50.
In simple words: First, we find the total cost formula by integrating the marginal cost. We use the fixed cost to find the unknown constant. Then, we put 4 into the total cost formula to find the cost of 4 units. For average cost, we divide the total cost formula by the number of units and then put 4 in again.
(ii) We are given the marginal cost function MC = \( 3(3x + 4)^{-1/2} \).
To find the total cost function C(x), we integrate MC(x):
\( C(x) = \int 3(3x + 4)^{-1/2} dx \)
Using the formula \( \int (ax+b)^n dx = \frac{(ax+b)^{n+1}}{a(n+1)} \):
\( C(x) = 3 \left[ \frac{(3x+4)^{(-1/2)+1}}{3((-1/2)+1)} \right] + K \)
\( C(x) = 3 \left[ \frac{(3x+4)^{1/2}}{3(1/2)} \right] + K \)
\( C(x) = 2\sqrt{3x+4} + K \) ...(1)
The fixed cost is 2. So, when \( x = 0, C(x) = 2 \).
Substitute these values into equation (1):
\( 2 = 2\sqrt{3(0)+4} + K \)
\( 2 = 2\sqrt{4} + K \)
\( 2 = 2(2) + K \)
\( 2 = 4 + K \)
\( K = 2 - 4 \implies K = -2 \)
Substitute the value of K back into equation (1) to get the total cost function:
\( C(x) = 2\sqrt{3x+4} - 2 \)
To find the average cost function (AC(x)), we divide the total cost function by x:
\( AC(x) = \frac{C(x)}{x} = \frac{2\sqrt{3x+4} - 2}{x} \)
To find the average cost of 7 units of output, we substitute \( x = 7 \) into AC(x):
\( AC(7) = \frac{2\sqrt{3(7)+4} - 2}{7} \)
\( AC(7) = \frac{2\sqrt{21+4} - 2}{7} \)
\( AC(7) = \frac{2\sqrt{25} - 2}{7} \)
\( AC(7) = \frac{2(5) - 2}{7} \)
\( AC(7) = \frac{10 - 2}{7} \)
\( AC(7) = \frac{8}{7} \)
The average cost of 7 units of output is \( \frac{8}{7} \).
In simple words: First, we find the total cost formula by integrating the marginal cost. We use the fixed cost to find the missing constant. Then, we find the average cost formula by dividing the total cost by 'x', and finally, we calculate the average cost for 7 units by plugging in \( x=7 \).

🎯 Exam Tip: When asked for the cost or average cost at a specific number of units, calculate the general C(x) or AC(x) first, and then substitute the specific value of x. Avoid direct substitution into the integral, which can lead to errors.

Question 9. Given that the marginal cost MC and the average cost AC of a product are directly proportional to each other, find the total cost function so that the cost of producing 2 units is 8 and of producing 4 units is 64.
Answer: We are given that MC is directly proportional to AC.
This means \( MC \propto AC \)
So, \( MC = K \cdot AC \) for some constant of proportionality K.
We know that \( MC = \frac{dC}{dx} \) and \( AC = \frac{C}{x} \).
Substituting these into the proportionality equation:
\( \frac{dC}{dx} = K \frac{C}{x} \)
Now, we separate the variables and integrate:
\( \frac{dC}{C} = K \frac{dx}{x} \)
\( \int \frac{dC}{C} = \int K \frac{dx}{x} \)
\( \log C = K \log x + \log K' \)
\( \log C = \log x^K + \log K' \)
\( \log C = \log (K' x^K) \)
This gives us the total cost function:
\( C(x) = K' x^K \) ...(1)
We are given that the cost of producing 2 units is 8. So, when \( x = 2, C(x) = 8 \).
Substitute into (1):
\( 8 = K' (2)^K \) ...(2)
We are also given that the cost of producing 4 units is 64. So, when \( x = 4, C(x) = 64 \).
Substitute into (1):
\( 64 = K' (4)^K \) ...(3)
Now, divide equation (3) by equation (2):
\( \frac{64}{8} = \frac{K' (4)^K}{K' (2)^K} \)
\( 8 = \left( \frac{4}{2} \right)^K \)
\( 8 = 2^K \)
Since \( 8 = 2^3 \):
\( 2^3 = 2^K \implies K = 3 \)
Now, substitute the value of K = 3 back into equation (2):
\( 8 = K' (2)^3 \)
\( 8 = K' (8) \)
\( K' = \frac{8}{8} \implies K' = 1 \)
Substitute the values of K and K' back into equation (1) to get the total cost function:
\( C(x) = 1 \cdot x^3 \)
\( C(x) = x^3 \)
The total cost function is \( C(x) = x^3 \).
In simple words: When marginal cost and average cost are linked in this way, we set up an equation using their definitions. By integrating and using the given costs for 2 and 4 units, we can find the values of the constants and determine the exact formula for the total cost.

🎯 Exam Tip: When given a proportionality relationship between MC and AC, remember to express MC as \( \frac{dC}{dx} \) and AC as \( \frac{C}{x} \). This leads to a separable differential equation, which can be solved by integration. Using the given data points helps determine the constants of integration and proportionality.