Samacheer Kalvi Class 12 Business Maths Solutions Chapter 3 Integral Calculus II Exercise 3.2

Question 1. The cost of an overhaul of an engine is Rs 10,000 The Operating cost per hour is at the rate of \( 2x-240 \) where the engine has run x km. find out the total cost of the engine run for 300 hours after overhaul.
Answer: Given that the overhaul cost is Rs. 10,000.
The marginal cost (MC) is \( 2x - 240 \).
The total cost (C) is found by integrating the marginal cost with respect to x, and adding a constant k:
\( C = \int MC \, dx + k \)
\( C = \int (2x - 240) \, dx + k \)
\( C = x^2 - 240x + k \)
Here, k represents the fixed overhaul cost.
Therefore, \( k = 10,000 \).
So, the total cost function is:
\( C = x^2 - 240x + 10,000 \)
Now, we need to find the total cost when the engine runs for 300 hours. So, substitute \( x = 300 \):
\( C = (300)^2 - 240(300) + 10,000 \)
\( C = 90,000 - 72,000 + 10,000 \)
\( C = 18,000 + 10,000 \)
\( C = 28,000 \)
Thus, the total cost for the engine run for 300 hours after the overhaul is Rs 28,000.
In simple words: First, we find the total cost formula by adding up the operating cost over time and the fixed overhaul cost. Then, we put in the number of hours (300) into this formula to get the total money spent.

🎯 Exam Tip: Remember that the constant of integration (k) in cost functions often represents the fixed cost or initial investment.

Question 2. Elasticity of a function \( \frac { Ey }{Ex} \) is given by \( \frac { Ey }{Ex} = \frac { -7x }{(1-2x)(2+3x)} \). Find the function when \( x = 2, y = \frac { 3 }{8} \).
Answer: We are given the elasticity of a function:
\( \frac { Ey }{Ex} = \frac { -7x }{(1-2x)(2+3x)} \)
We know that elasticity \( \frac { Ey }{Ex} = \frac { x }{y} \frac { dy }{dx} \).
So, \( \frac { x }{y} \frac { dy }{dx} = \frac { -7x }{(1-2x)(2+3x)} \)
Divide both sides by x (assuming \( x \neq 0 \)):
\( \frac { 1 }{y} \frac { dy }{dx} = \frac { -7 }{(1-2x)(2+3x)} \)
To solve this, we separate the variables:
\( \frac { 1 }{y} dy = \frac { -7 }{(1-2x)(2+3x)} dx \)
We can rewrite the right side by factoring out -1 from (1-2x) to make it (2x-1):
\( \frac { 1 }{y} dy = \frac { 7 }{(2x-1)(2+3x)} dx \) ....(1)
Now, we use partial fraction decomposition for the right side:
\( \frac { 7 }{(2x-1)(3x+2)} = \frac { A }{2x-1} + \frac { B }{3x+2} \)
\( 7 = A(3x+2) + B(2x-1) \)
To find A, set \( 2x-1 = 0 \implies x = \frac { 1 }{2} \):
\( 7 = A(3(\frac { 1 }{2})+2) + B(2(\frac { 1 }{2})-1) \)
\( 7 = A(\frac { 3 }{2}+2) + B(1-1) \)
\( 7 = A(\frac { 3+4 }{2}) + B(0) \)
\( 7 = A(\frac { 7 }{2}) \)
\( \implies A = 2 \)
To find B, set \( 3x+2 = 0 \implies x = -\frac { 2 }{3} \):
\( 7 = A(3(-\frac { 2 }{3})+2) + B(2(-\frac { 2 }{3})-1) \)
\( 7 = A(-2+2) + B(-\frac { 4 }{3}-1) \)
\( 7 = A(0) + B(-\frac { 4+3 }{3}) \)
\( 7 = B(-\frac { 7 }{3}) \)
\( \implies B = -3 \)
So, equation (1) becomes:
\( \frac { 1 }{y} dy = (\frac { 2 }{2x-1} - \frac { 3 }{3x+2}) dx \)
Now, integrate both sides:
\( \int \frac { 1 }{y} dy = \int (\frac { 2 }{2x-1} - \frac { 3 }{3x+2}) dx \)
\( \log |y| = 2 \cdot \frac { \log |2x-1| }{2} - 3 \cdot \frac { \log |3x+2| }{3} + \log k \)
\( \log |y| = \log |2x-1| - \log |3x+2| + \log k \)
\( \log |y| = \log \left| \frac { k(2x-1) }{3x+2} \right| \)
This gives us the function:
\( y = \frac { k(2x-1) }{3x+2} \) ....(2)
We are given that when \( x = 2 \), \( y = \frac { 3 }{8} \). Substitute these values to find k:
\( \frac { 3 }{8} = \frac { k(2(2)-1) }{3(2)+2} \)
\( \frac { 3 }{8} = \frac { k(3) }{8} \)
Multiply both sides by 8:
\( 3 = 3k \)
\( \implies k = 1 \)
Substitute \( k=1 \) back into equation (2):
\( y = \frac { 1(2x-1) }{3x+2} \)
Thus, the function is \( y = \frac { 2x-1 }{3x+2} \).
In simple words: We used the given elasticity formula and changed it to a form we could integrate. We broke down the right side into simpler fractions to make integration easier. After integrating, we used the given point (x=2, y=3/8) to find the missing constant, giving us the final function.

🎯 Exam Tip: Remember the formula for elasticity and be careful with partial fraction decomposition. Always substitute the initial conditions at the end to find the constant of integration.

Question 3. The Elasticity of demand with respect to price for a commodity is given by where \( \frac { (4-x)}{x} \) p is the price when demand is x. find the demand function when the price is 4 and the demand is 2. Also, find the revenue function
Answer: The elasticity of demand (\( \eta_d \)) with respect to price is given as:
\( \eta_d = \frac { -(4-x)}{x} \)
We know the formula for elasticity of demand:
\( \eta_d = \frac { -p }{x} \frac { dx }{dp} \)
So, \( \frac { -p }{x} \frac { dx }{dp} = \frac { -(4-x)}{x} \)
Multiply both sides by \( \frac { -x }{p} \):
\( \frac { dx }{dp} = \frac { (4-x)}{p} \)
Now, we separate the variables to integrate:
\( \frac { dx }{4-x} = \frac { dp }{p} \)
Integrate both sides:
\( \int \frac { 1 }{4-x} dx = \int \frac { 1 }{p} dp \)
\( - \log |4-x| = \log |p| + \log k \)
\( \log |(4-x)^{-1}| = \log |pk| \)
\( \log \left| \frac { 1 }{4-x} \right| = \log |pk| \)
This means:
\( \frac { 1 }{4-x} = pk \)
\( (4-x)pk = 1 \)
\( (4-x) = \frac { 1 }{pk} \) ....(1)
We are given that when price \( p = 4 \), demand \( x = 2 \). Substitute these values into the equation to find k:
\( (4-2) = \frac { 1 }{k(4)} \)
\( 2 = \frac { 1 }{4k} \)
\( 8k = 1 \)
\( \implies k = \frac { 1 }{8} \)
Substitute \( k = \frac { 1 }{8} \) back into equation (1):
\( (4-x) = \frac { 1 }{p(\frac { 1 }{8})} \)
\( (4-x) = \frac { 8 }{p} \)
This is the demand function. We can also express p in terms of x:
\( p(4-x) = 8 \)
\( p = \frac { 8 }{4-x} \)
Next, we need to find the revenue function (R). Revenue is given by \( R = px \).
Substitute the expression for p:
\( R = \left( \frac { 8 }{4-x} \right) x \)
\( R = \frac { 8x }{4-x} \)
In simple words: We used the given elasticity formula and related it to price and demand. By doing some math, including integration, we found a relationship between price and demand. Then, we used the given example (price 4, demand 2) to find a missing number. Finally, we used this to get the demand function and the revenue function.

🎯 Exam Tip: Pay close attention to the sign conventions for elasticity of demand and remember that the integration constant 'k' can be determined using the given specific price-demand pair.

Question 4. A company receives a shipment of 500 scooters every 30 days. From experience it is known that the inventory on hand is related to the number of days x. Since the shipment, I (x) = \( 500 - 0.03 x^2 \), the daily holding cost per scooter is Rs 0.3. Determine the total cost for maintaining inventory for 30 days
Answer: Given information:
Inventory on hand, \( I(x) = 500 - 0.03 x^2 \)
Daily holding cost per scooter, \( C_1 = Rs \ 0.3 \)
Time period, \( T = 30 \) days
The total inventory carrying cost is calculated by integrating the product of the daily holding cost and the inventory function over the given time period.
Total inventory carrying cost \( = \int_{0}^{T} C_1 \cdot I(x) \, dx \)
Total inventory carrying cost \( = \int_{0}^{30} 0.3 (500 - 0.03 x^2) \, dx \)
\( = 0.3 \int_{0}^{30} (500 - 0.03 x^2) \, dx \)
\( = 0.3 \left[ 500x - 0.03 \frac {x^3 }{3} \right]_{0}^{30} \)
\( = 0.3 \left[ 500x - 0.01 x^3 \right]_{0}^{30} \)
Now, substitute the limits of integration:
\( = 0.3 \left( (500(30) - 0.01 (30)^3) - (500(0) - 0.01 (0)^3) \right) \)
\( = 0.3 \left( (15000 - 0.01 (27000)) - (0) \right) \)
\( = 0.3 (15000 - 270) \)
\( = 0.3 (14730) \)
\( = 4419 \)
The total cost for maintaining inventory for 30 days is Rs 4,419.
In simple words: To find the total cost of keeping scooters in stock, we multiply the daily cost per scooter by how many scooters are there each day and add it all up for 30 days using integration. We then put in the numbers for 30 days and calculate the final amount.

🎯 Exam Tip: For inventory problems, clearly identify the inventory function and the holding cost. Remember to set up and evaluate the definite integral correctly for the given time period.

Question 5. An account fetches interest at the rate of 5% per annum compounded continuously an individual deposits Rs 1000 each year in his account. how much will be in the account after 5 years ( \( e^{0.25} = 1.284 \) )
Answer: Given information:
Annual deposit (P) = Rs 1000
Interest rate (r) = 5% per annum compounded continuously \( = \frac { 5 }{100} = 0.05 \)
Time period (N) = 5 years
When interest is compounded continuously and deposits are made annually, we use the formula for future value of an annuity with continuous compounding:
Annuity \( = \int_{0}^{N} P e^{rt} dt \)
\( = \int_{0}^{5} 1000 e^{0.05t} dt \)
\( = 1000 \left[ \frac { e^{0.05t} }{0.05} \right]_{0}^{5} \)
\( = \frac { 1000 }{0.05} [e^{0.05 \times 5} - e^{0.05 \times 0}] \)
\( = 20000 [e^{0.25} - e^0] \)
Since \( e^0 = 1 \), and given \( e^{0.25} = 1.284 \):
\( = 20000 [1.284 - 1] \)
\( = 20000 [0.284] \)
\( = 5680 \)
After 5 years, the total amount in the account will be Rs 5680.
In simple words: This problem asks for the total money in an account after 5 years if someone puts in Rs 1000 every year, and the bank adds interest continuously. We use a special formula that involves integration to calculate how all these deposits grow over time with continuous interest.

🎯 Exam Tip: For continuous compounding and annuities, remember the integration formula \( \int P e^{rt} dt \). Pay attention to the limits of integration and the given value of e to a power.

Question 6. The marginal cost function of a product is given by \( \frac { dc }{dx} = 100 – 10x + 0.1 x^2 \) where x is the output. Obtain the total and average cost function of the firm under the assumption, that its fixed cost is Rs 500
Answer: Given the marginal cost function:
\( \frac { dc }{dx} = 100 - 10x + 0.1 x^2 \)
To find the total cost function (C), we integrate the marginal cost function:
\( C = \int \frac { dc }{dx} \, dx \)
\( C = \int (100 - 10x + 0.1 x^2) \, dx \)
\( C = 100x - 10\frac {x^2 }{2} + 0.1\frac {x^3 }{3} + k \)
\( C = 100x - 5x^2 + \frac {0.1}{3} x^3 + k \)
We are given that the fixed cost is Rs 500. Fixed cost is the cost when output \( x = 0 \). So, \( C(0) = 500 \).
Substitute \( x = 0 \) into the total cost function:
\( 500 = 100(0) - 5(0)^2 + \frac {0.1}{3} (0)^3 + k \)
\( \implies k = 500 \)
So, the total cost function is:
\( C = 100x - 5x^2 + \frac {0.1}{3} x^3 + 500 \)
We can also write \( \frac {0.1}{3} \) as \( \frac {1}{30} \).
\( C = 100x - 5x^2 + \frac {x^3 }{30} + 500 \)
Now, to find the average cost function (AC), we divide the total cost function by x:
\( AC = \frac { C }{x} \)
\( AC = \frac { 100x - 5x^2 + \frac {x^3 }{30} + 500 }{x} \)
\( AC = 100 - 5x + \frac {x^2 }{30} + \frac { 500 }{x} \)
In simple words: We started with the cost to make one extra item (marginal cost). To get the total cost, we added up all these small costs, which means using integration. We then used the given fixed cost to find a missing number in our total cost formula. Finally, we divided the total cost by the number of items to get the average cost for each item.

🎯 Exam Tip: Remember that total cost is the integral of marginal cost. The constant of integration (k) for cost functions always represents the fixed cost when output is zero.

Question 7. The marginal cost function is M.C = \( 300 x^{2/5} \) and the fixed cost is zero. Find the total cost as a function of x
Answer: Given the marginal cost function:
\( MC = 300 x^{2/5} \)
The fixed cost is zero, which means the constant of integration will be zero.
To find the total cost (C) as a function of x, we integrate the marginal cost function:
\( C = \int MC \, dx \)
\( C = \int 300 x^{2/5} \, dx \)
\( C = 300 \left( \frac { x^{\frac { 2 }{5} + 1} }{\frac { 2 }{5} + 1} \right) + k \)
\( C = 300 \left( \frac { x^{\frac { 7 }{5}} }{\frac { 7 }{5}} \right) + k \)
\( C = 300 \cdot \frac { 5 }{7} x^{7/5} + k \)
\( C = \frac { 1500 }{7} x^{7/5} + k \)
Since the fixed cost is zero, \( k = 0 \).
So, the total cost function is:
\( C = \frac { 1500 }{7} x^{7/5} \)
If we were asked for the average cost (AC), it would be \( AC = \frac { C }{x} = \frac { 1500 }{7} x^{7/5-1} = \frac { 1500 }{7} x^{2/5} \).
In simple words: We are given the cost to make one more item (marginal cost) and that there are no fixed costs. To find the total cost for making many items, we add up all the small marginal costs using integration. Since there are no fixed costs, we don't add any extra constant at the end.

🎯 Exam Tip: When integrating \( x^n \), remember the power rule: \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \). Be careful with fractional exponents, and always correctly apply the constant of integration, 'k', which is 0 if fixed cost is zero.

Question 8. If the marginal cost function of x units of output is \( \frac { a }{\sqrt {ax+b}} \) and if the cost of output is zero. Find the total cost as a function of x.
Answer: Given the marginal cost function (MC):
\( MC = \frac { a }{\sqrt {ax+b}} \)
We can rewrite this as: \( MC = a(ax+b)^{-1/2} \)
To find the total cost function (C), we integrate the marginal cost function:
\( C = \int MC \, dx \)
\( C = \int a(ax+b)^{-1/2} \, dx \)
To integrate this, we use the substitution method or the general power rule for integration: \( \int (cx+d)^n dx = \frac{(cx+d)^{n+1}}{c(n+1)} \).
Here, \( c=a \), \( d=b \), \( n = -\frac { 1 }{2} \).
\( C = a \left( \frac { (ax+b)^{-1/2+1} }{a(-\frac { 1 }{2}+1)} \right) + k \)
\( C = a \left( \frac { (ax+b)^{1/2} }{a(\frac { 1 }{2})} \right) + k \)
\( C = \frac { a(ax+b)^{1/2} }{\frac { a }{2}} + k \)
\( C = 2(ax+b)^{1/2} + k \)
So, the total cost function is \( C(x) = 2\sqrt {ax+b} + k \).
We are given that the cost of output is zero when \( x = 0 \). This means \( C(0) = 0 \).
Substitute \( x = 0 \) into the cost function:
\( 0 = 2\sqrt {a(0)+b} + k \)
\( 0 = 2\sqrt {b} + k \)
\( \implies k = -2\sqrt {b} \)
Now substitute the value of k back into the total cost function:
\( C(x) = 2\sqrt {ax+b} - 2\sqrt {b} \)
This is the total cost as a function of x.
In simple words: We started with the formula for marginal cost (how much it costs to make one more item). To find the total cost for all items, we used a math method called integration. Then, we used the fact that there's no cost when nothing is produced to find a missing number in our total cost formula, giving us the final answer.

🎯 Exam Tip: When integrating functions involving a linear term inside a power or root (like \( (ax+b)^n \)), remember to divide by 'a' (the coefficient of x) as part of the integration rule.

Question 9. Determine the cost of producing 200 air conditioners if the marginal cost (is per unit) is \( C'(x) = \frac { x^2 }{200} + 4 \)
Answer: Given the marginal cost function:
\( C'(x) = \frac { x^2 }{200} + 4 \)
To find the total cost function (C), we integrate the marginal cost function:
\( C = \int C'(x) \, dx \)
\( C = \int \left( \frac { x^2 }{200} + 4 \right) \, dx \)
\( C = \frac { 1 }{200} \frac { x^3 }{3} + 4x + k \)
\( C = \frac { x^3 }{600} + 4x + k \)
We need an initial condition to find k. The problem doesn't state a fixed cost, but in such cases, it's often implied that fixed cost is 0, meaning when \( x = 0 \), \( C = 0 \).
Let's assume fixed cost is zero, so \( C(0) = 0 \):
\( 0 = \frac { (0)^3 }{600} + 4(0) + k \)
\( \implies k = 0 \)
So, the total cost function is:
\( C(x) = \frac { x^3 }{600} + 4x \)
Now, we need to determine the cost of producing 200 air conditioners. Substitute \( x = 200 \) into the cost function:
\( C(200) = \frac { (200)^3 }{600} + 4(200) \)
\( C(200) = \frac { 8000000 }{600} + 800 \)
\( C(200) = 13333.333... + 800 \)
\( C(200) = 14133.33 \)
The cost of producing 200 air conditioners is approximately Rs 14133.33.
In simple words: We used the marginal cost (cost to make one more air conditioner) to find a formula for the total cost of making any number of air conditioners, using integration. We assumed there are no fixed costs. Then, we put in the number 200 into our total cost formula to find out how much it would cost to make 200 air conditioners.

🎯 Exam Tip: Always remember to consider fixed costs (the constant of integration, k) in total cost functions. If not explicitly given, assume it to be zero unless context implies otherwise.

Question 10. The marginal revenue (in thousands of Rupees) function for a particular commodity is \( 5 + 3 e^{-0.03x} \) where x denotes the number of units sold. Determine the total revenue from the sale of 100 units (given \( e^{-3} = \) approximately)
Answer: Given the marginal revenue (MR) function in thousands of Rupees:
\( MR = 5 + 3 e^{-0.03x} \)
To find the total revenue (TR) from the sale of 100 units, we integrate the marginal revenue function from 0 to 100:
\( TR = \int_{0}^{100} MR \, dx \)
\( TR = \int_{0}^{100} (5 + 3 e^{-0.03x}) \, dx \)
\( TR = \left[ 5x + 3 \frac { e^{-0.03x} }{-0.03} \right]_{0}^{100} \)
\( TR = \left[ 5x - \frac { 3 }{0.03} e^{-0.03x} \right]_{0}^{100} \)
\( TR = \left[ 5x - 100 e^{-0.03x} \right]_{0}^{100} \)
Now, we evaluate the expression at the limits:
\( TR = (5(100) - 100 e^{-0.03 \times 100}) - (5(0) - 100 e^{-0.03 \times 0}) \)
\( TR = (500 - 100 e^{-3}) - (0 - 100 e^0) \)
Given \( e^{-3} \approx 0.05 \) and \( e^0 = 1 \):
\( TR = (500 - 100(0.05)) - (0 - 100(1)) \)
\( TR = (500 - 5) - (-100) \)
\( TR = 495 + 100 \)
\( TR = 595 \)
Since the revenue is in thousands of Rupees, the total revenue is \( 595 \times 1000 = Rs \ 595000 \).
In simple words: We have a formula for how much extra money you get from selling one more item (marginal revenue). To find the total money from selling 100 items, we added up all these small amounts using integration. Then, we used the given values to calculate the total amount of money earned.

🎯 Exam Tip: When integrating exponential functions \( e^{ax} \), remember that \( \int e^{ax} dx = \frac{1}{a} e^{ax} \). Pay close attention to the negative sign in the exponent and evaluating the function at both limits correctly.

Question 11. If the marginal revenue function for a commodity is MR = \( 9 – 4x^2 \). Find the demand function.
Answer: Given the marginal revenue (MR) function:
\( MR = 9 - 4x^2 \)
To find the total revenue function (R), we integrate the marginal revenue function:
\( R = \int MR \, dx \)
\( R = \int (9 - 4x^2) \, dx \)
\( R = 9x - 4\frac {x^3 }{3} + k \)
Typically, when \( x = 0 \) (no units sold), total revenue \( R = 0 \). So, we can assume \( k = 0 \).
\( R = 9x - \frac {4x^3 }{3} \)
The demand function (p) is defined as average revenue, which is total revenue divided by the number of units (x):
\( p = \frac { R }{x} \)
\( p = \frac { 9x - \frac {4x^3 }{3} }{x} \)
\( p = \frac { 9x }{x} - \frac { \frac {4x^3 }{3} }{x} \)
\( p = 9 - \frac {4x^2 }{3} \)
This is the demand function.
In simple words: We started with the extra money earned from selling one more item (marginal revenue). We integrated this to find the total money earned from selling all items. Then, to get the demand function, which shows the price per item, we divided the total money earned by the number of items sold.

🎯 Exam Tip: Remember the relationship between marginal revenue, total revenue, and demand function: \( R = \int MR \, dx \) and \( p = \frac{R}{x} \). Assume the constant of integration for total revenue is zero if no fixed revenue is specified.

Question 12. Given marginal revenue function \( \frac { 4 }{(2x+3)^2} -1 \), show that the average revenue function is \( P = \frac { 4 }{6x+9} -1 \)
Answer: Given the marginal revenue (MR) function:
\( MR = \frac { 4 }{(2x+3)^2} - 1 \)
We can write this as: \( MR = 4(2x+3)^{-2} - 1 \)
To find the total revenue function (R), we integrate the marginal revenue function:
\( R = \int MR \, dx \)
\( R = \int (4(2x+3)^{-2} - 1) \, dx \)
For the first term, use the power rule for integration: \( \int (cx+d)^n dx = \frac{(cx+d)^{n+1}}{c(n+1)} \). Here \( c=2, n=-2 \).
\( R = 4 \frac { (2x+3)^{-2+1} }{2(-2+1)} - x + k \)
\( R = 4 \frac { (2x+3)^{-1} }{2(-1)} - x + k \)
\( R = 4 \frac { (2x+3)^{-1} }{-2} - x + k \)
\( R = -2(2x+3)^{-1} - x + k \)
\( R = \frac { -2 }{2x+3} - x + k \)
Assuming total revenue is zero when \( x = 0 \) (no sales), we find k:
\( 0 = \frac { -2 }{2(0)+3} - 0 + k \)
\( 0 = \frac { -2 }{3} + k \)
\( \implies k = \frac { 2 }{3} \)
So, the total revenue function is:
\( R = \frac { -2 }{2x+3} - x + \frac { 2 }{3} \)
Now, we need to find the average revenue (AR), which is \( P = \frac { R }{x} \):
\( P = \frac { 1 }{x} \left( \frac { -2 }{2x+3} - x + \frac { 2 }{3} \right) \)
\( P = \frac { -2 }{x(2x+3)} - \frac { x }{x} + \frac { 2 }{3x} \)
\( P = \frac { -2 }{2x^2+3x} - 1 + \frac { 2 }{3x} \)
This form doesn't immediately match \( \frac { 4 }{6x+9} -1 \). Let's re-examine the given average revenue. It can be written as \( \frac { 4 }{3(2x+3)} -1 \).
Let's adjust our total revenue to match the desired average revenue:
\( R = \frac { 4x }{6x+9} - x \)
\( R = \frac { 4x }{3(2x+3)} - x \)
If this is the form of R, then
\( \frac { -2 }{2x+3} - x + \frac { 2 }{3} \) must be equal to \( \frac { 4x }{3(2x+3)} - x \).
\( \frac { -2 }{2x+3} + \frac { 2 }{3} = \frac { 4x }{3(2x+3)} \)
\( \frac { -2(3) + 2(2x+3) }{3(2x+3)} = \frac { 4x }{3(2x+3)} \)
\( -6 + 4x + 6 = 4x \)
\( 4x = 4x \)
This identity shows that our total revenue function (when k is determined) is consistent with the desired average revenue function. So, if we take \( R = \frac { 4x }{3(2x+3)} \), then \( AR = \frac { R }{x} = \frac { 4 }{3(2x+3)} = \frac { 4 }{6x+9} \). The -1 term must have been part of the initial MR.
Let's consider the problem statement more carefully. The question implies that the marginal revenue function given leads to the average revenue function *after integration*. The "-1" in MR is a constant, and it should carry over.
Given \( MR = \frac { 4 }{(2x+3)^2} -1 \).
\( R = \int (\frac { 4 }{(2x+3)^2} -1) \, dx \)
\( R = \frac { -2 }{2x+3} - x + k \). We found \( k = \frac { 2 }{3} \).
\( R = \frac { -2 }{2x+3} - x + \frac { 2 }{3} \)
Then, average revenue \( P = \frac { R }{x} = \frac { 1 }{x} \left( \frac { -2 }{2x+3} - x + \frac { 2 }{3} \right) \)
\( P = \frac { -2 }{x(2x+3)} - 1 + \frac { 2 }{3x} \)
This does not directly simplify to \( \frac { 4 }{6x+9} -1 \).
Let's consider an alternative way the given average revenue might arise. If \( P = \frac { 4 }{6x+9} -1 \), then \( R = Px = \frac { 4x }{6x+9} - x \).
Then \( MR = \frac { dR }{dx} \).
Let \( R = \frac { 4x }{3(2x+3)} - x \).
Using quotient rule for \( \frac { 4x }{3(2x+3)} \): \( \frac { d }{dx} \left( \frac { 4x }{3(2x+3)} \right) = \frac { 3(2x+3)(4) - 4x(3)(2) }{[3(2x+3)]^2} = \frac { 12(2x+3) - 24x }{9(2x+3)^2} = \frac { 24x+36-24x }{9(2x+3)^2} = \frac { 36 }{9(2x+3)^2} = \frac { 4 }{(2x+3)^2} \).
And \( \frac { d }{dx}(-x) = -1 \).
So, \( MR = \frac { 4 }{(2x+3)^2} - 1 \). This matches the given MR.
This means the given average revenue \( P = \frac { 4 }{6x+9} -1 \) is correct. The working should lead to this.
Let's re-evaluate total revenue \( R \). It is known that \( R = \frac { -2 }{2x+3} - x + k \).
If \( P = \frac { 4 }{6x+9} -1 \), then it must be that \( R = x(\frac { 4 }{6x+9} -1) = \frac { 4x }{6x+9} - x \).
So, \( \frac { -2 }{2x+3} - x + k = \frac { 4x }{6x+9} - x \)
\( \frac { -2 }{2x+3} + k = \frac { 4x }{3(2x+3)} \)
If this is true for all x, then \( k = \frac { 4x }{3(2x+3)} + \frac { 2 }{2x+3} = \frac { 4x + 2(3) }{3(2x+3)} = \frac { 4x+6 }{3(2x+3)} = \frac { 2(2x+3) }{3(2x+3)} = \frac { 2 }{3} \).
This confirms that \( k = \frac { 2 }{3} \) is correct. So, the total revenue function is indeed \( R = \frac { -2 }{2x+3} - x + \frac { 2 }{3} \).
Now, we need to manipulate \( P = \frac { R }{x} \) to match \( \frac { 4 }{6x+9} -1 \).
\( P = \frac { 1 }{x} \left( \frac { -2 }{2x+3} - x + \frac { 2 }{3} \right) \)
\( P = \frac { -2 }{x(2x+3)} - 1 + \frac { 2 }{3x} \)
To get the target form \( \frac { 4 }{3(2x+3)} - 1 \), we need to simplify \( \frac { -2 }{x(2x+3)} + \frac { 2 }{3x} \):
\( \frac { -2 \cdot 3 + 2(2x+3) }{3x(2x+3)} = \frac { -6 + 4x + 6 }{3x(2x+3)} = \frac { 4x }{3x(2x+3)} = \frac { 4 }{3(2x+3)} \)
Thus, \( P = \frac { 4 }{3(2x+3)} - 1 \).
\( P = \frac { 4 }{6x+9} - 1 \). This shows the desired result.
In simple words: We started with the formula for how much extra money you get from selling one more item. We used integration to find the total money earned. Then, we divided the total money by the number of items sold to get the average money per item (average revenue). We made sure our answer matched the target formula by carefully combining the terms.

🎯 Exam Tip: When showing that a derived function matches a target, it's helpful to either work from MR to AR, or differentiate the target AR (after multiplying by x to get R) to see if it matches the original MR. Careful algebraic manipulation is key.

Question 13. A firms marginal revenue functions is M.R = \( 20 e^{-x/10} \) Find the corresponding demand function.
Answer: Given the marginal revenue (MR) function:
\( MR = 20 e^{-x/10} \)
To find the total revenue function (R), we integrate the marginal revenue function:
\( R = \int MR \, dx \)
\( R = \int 20 e^{-x/10} \, dx \)
We use the integration rule \( \int e^{ax} dx = \frac{1}{a} e^{ax} \). Here \( a = -\frac { 1 }{10} \).
\( R = 20 \left( \frac { e^{-x/10} }{-\frac { 1 }{10}} \right) + k \)
\( R = 20 (-10) e^{-x/10} + k \)
\( R = -200 e^{-x/10} + k \)
Typically, when \( x = 0 \) (no units sold), total revenue \( R = 0 \). Substitute these values to find k:
\( 0 = -200 e^{-0/10} + k \)
\( 0 = -200 e^0 + k \)
\( 0 = -200(1) + k \)
\( \implies k = 200 \)
So, the total revenue function is:
\( R = -200 e^{-x/10} + 200 \)
\( R = 200(1 - e^{-x/10}) \)
The demand function (p) is the average revenue, which is total revenue divided by the number of units (x):
\( p = \frac { R }{x} \)
\( p = \frac { 200(1 - e^{-x/10}) }{x} \)
This is the demand function.
In simple words: We used the marginal revenue (the extra money from selling one more item) to find the total money earned from selling all items using integration. We found a constant by assuming no revenue from zero sales. Then, we divided the total revenue by the number of items to find the demand function, which tells us the price per item.

🎯 Exam Tip: For exponential functions like \( e^{ax} \), remember the integration rule \( \int e^{ax} dx = \frac{1}{a} e^{ax} \). Always find the constant of integration using the condition that revenue is zero when sales are zero.

Question 14. The marginal cost of production of a firm is given by C' (x) = \( 5 + 0.13x \), the marginal revenue is given by R' (x) = 18 and the fixed cost is Rs 120. Find the profit function.
Answer: Given information:
Marginal cost function, \( C'(x) = 5 + 0.13x \)
Marginal revenue function, \( R'(x) = 18 \)
Fixed cost = Rs 120
First, find the total cost function (C(x)) by integrating \( C'(x) \):
\( C(x) = \int (5 + 0.13x) \, dx \)
\( C(x) = 5x + 0.13 \frac {x^2 }{2} + k_1 \)
\( C(x) = 5x + 0.065x^2 + k_1 \)
We know the fixed cost is Rs 120. Fixed cost is \( C(0) \).
So, when \( x = 0 \), \( C(0) = 120 \).
\( 120 = 5(0) + 0.065(0)^2 + k_1 \)
\( \implies k_1 = 120 \)
The total cost function is: \( C(x) = 5x + 0.065x^2 + 120 \)
Next, find the total revenue function (R(x)) by integrating \( R'(x) \):
\( R(x) = \int 18 \, dx \)
\( R(x) = 18x + k_2 \)
Assuming revenue is zero when \( x = 0 \), then \( k_2 = 0 \).
The total revenue function is: \( R(x) = 18x \)
Finally, the profit function (P(x)) is given by Total Revenue - Total Cost:
\( P(x) = R(x) - C(x) \)
\( P(x) = 18x - (5x + 0.065x^2 + 120) \)
\( P(x) = 18x - 5x - 0.065x^2 - 120 \)
\( P(x) = 13x - 0.065x^2 - 120 \)
This is the profit function.
In simple words: We were given the cost to make one extra item and the money made from selling one extra item. We used these to find the total cost and total money earned using integration. We included the fixed cost in the total cost. Then, we subtracted the total cost from the total money earned to find the overall profit formula.

🎯 Exam Tip: Remember that profit is always Total Revenue minus Total Cost. Be careful to correctly determine the constants of integration for both cost (fixed cost) and revenue (usually zero at zero output).

Question 15. If the marginal revenue function is R'(x) = \( 1500 – 4x – 3x^2 \). Find the revenue function and average revenue function.
Answer: Given the marginal revenue (MR) function:
\( R'(x) = 1500 - 4x - 3x^2 \)
To find the total revenue function (R), we integrate the marginal revenue function:
\( R(x) = \int R'(x) \, dx \)
\( R(x) = \int (1500 - 4x - 3x^2) \, dx \)
\( R(x) = 1500x - 4\frac {x^2 }{2} - 3\frac {x^3 }{3} + c \)
\( R(x) = 1500x - 2x^2 - x^3 + c \)
Assuming revenue is zero when \( x = 0 \) (no units sold), then \( c = 0 \).
So, the total revenue function is:
\( R(x) = 1500x - 2x^2 - x^3 \)
Next, to find the average revenue function (P), we divide the total revenue function by x:
\( P = \frac { R(x) }{x} \)
\( P = \frac { 1500x - 2x^2 - x^3 }{x} \)
\( P = \frac { 1500x }{x} - \frac { 2x^2 }{x} - \frac { x^3 }{x} \)
\( P = 1500 - 2x - x^2 \)
This is the average revenue function.
In simple words: We used the given formula for how much extra money you get from selling one more item (marginal revenue). We integrated this to find the total money earned from selling all items. Then, we divided the total money earned by the number of items sold to find the average money made per item.

🎯 Exam Tip: Always remember that Total Revenue is the integral of Marginal Revenue, and Average Revenue is Total Revenue divided by the quantity (x). The constant of integration for revenue is typically zero if no revenue exists at zero output.

Question 17. The marginal cost function of a commodity is given by Mc = \( \frac { 14000 }{\sqrt{7x+4}} \) and the fixed cost is Rs 18,000. Find the total cost average cost.
Answer:
Given marginal cost function \( M_c = \frac{14000}{\sqrt{7x+4}} = 14000 (7x+4)^{-1/2} \).
The fixed cost is \( k = \text{Rs } 18,000 \).
To find the total cost function (C), we integrate the marginal cost function:
\( C = \int M_c \,dx \)
\( C = \int 14000 (7x+4)^{-1/2} \,dx \)
Now, we apply the power rule for integration.
\( C = 14000 \left[ \frac{(7x+4)^{-1/2+1}}{(-1/2+1) \times 7} \right] + k \)
\( C = 14000 \left[ \frac{(7x+4)^{1/2}}{(1/2) \times 7} \right] + k \)
\( C = 14000 \left[ \frac{2 \sqrt{7x+4}}{7} \right] + k \)
\( C = 2000 \times 2 \sqrt{7x+4} + k \)
\( C = 4000 \sqrt{7x+4} + k \)
Substitute the fixed cost value for k:
\( C = 4000 \sqrt{7x+4} + 18000 \)
The total cost function is now complete.
To find the average cost function (A.C), we divide the total cost by the number of units (x):
\( A.C = \frac{C}{x} \)
\( A.C = \frac{4000 \sqrt{7x+4} + 18000}{x} \)
\( A.C = \frac{4000 \sqrt{7x+4}}{x} + \frac{18000}{x} \)
In simple words: We find the total cost by adding up all the small marginal costs, which means using integration. Then, we find the average cost by dividing the total cost by how many items were produced.

🎯 Exam Tip: Remember that fixed cost is represented by the constant of integration (k) when finding the total cost from the marginal cost function.

Question 18. If the marginal cost (MC) of production of the company is directly proportional to the number of units (x) produced, then find the total cost function, when the fixed cost is Rs 5,000 and the cost of producing 50 units is Rs 5,625.
Answer:
Given that marginal cost (MC) is directly proportional to the number of units (x), we can write:
\( M_C = \lambda x \)
Here, \( \lambda \) is the constant of proportionality.
The fixed cost is given as \( k = \text{Rs } 5000 \).
The cost of producing 50 units is \( C(50) = \text{Rs } 5625 \).
To find the total cost function (C), we integrate the marginal cost function:
\( C = \int M_C \,dx = \int \lambda x \,dx \)
\( C = \frac{\lambda x^2}{2} + k \)
Substitute the fixed cost k into the equation:
\( C = \frac{\lambda x^2}{2} + 5000 \)
Now, use the given information that \( C(50) = 5625 \) to find the value of \( \lambda \):
\( 5625 = \frac{\lambda (50)^2}{2} + 5000 \)
\( 5625 - 5000 = \frac{\lambda \times 2500}{2} \)
\( 625 = 1250 \lambda \)
\( \implies \lambda = \frac{625}{1250} \)
\( \implies \lambda = \frac{1}{2} \)
Finally, substitute the value of \( \lambda \) back into the total cost function:
\( C = \frac{1}{2} \frac{x^2}{2} + 5000 \)
\( C = \frac{x^2}{4} + 5000 \)
In simple words: We found a formula for the total cost. First, we used the idea that cost changes with how many items are made. Then, we used the given fixed cost and the cost for 50 items to figure out all the missing numbers in our formula.

🎯 Exam Tip: Always remember that "directly proportional" means setting up an equation like \( y = kx \), and the fixed cost is the constant of integration when working from marginal cost to total cost.

Question 19. If MR = 20 – 5x + 3x², Find total revenue function
Answer:
Given the marginal revenue function \( M_R = 20 - 5x + 3x^2 \).
To find the total revenue function (R), we integrate the marginal revenue function:
\( R = \int M_R \,dx = \int (20 - 5x + 3x^2) \,dx \)
We apply the rules of integration for each term:
\( R = 20x - \frac{5x^2}{2} + \frac{3x^3}{3} + k \)
Simplify the expression:
\( R = 20x - \frac{5x^2}{2} + x^3 + k \)
In most cases, when the output (x) is zero, the total revenue (R) is also zero. This helps us find the constant of integration (k).
If \( x = 0 \), then \( R = 0 \):
\( 0 = 20(0) - \frac{5(0)^2}{2} + (0)^3 + k \)
\( \implies k = 0 \)
So, the total revenue function is:
\( R = 20x - \frac{5x^2}{2} + x^3 \)
In simple words: We started with how much extra money each new item brings in (marginal revenue). By adding all these small amounts together, we found the total money earned from selling all items. We assume no sales means no money earned.

🎯 Exam Tip: For revenue functions, the constant of integration (k) is typically assumed to be zero, as there is no revenue when zero units are sold.

Question 20. If MR = 14 – 6x + 9x², Find the demand function.
Answer:
Given the marginal revenue function \( M_R = 14 - 6x + 9x^2 \).
First, we need to find the total revenue function (R) by integrating the marginal revenue function:
\( R = \int M_R \,dx = \int (14 - 6x + 9x^2) \,dx \)
Integrate each term:
\( R = 14x - \frac{6x^2}{2} + \frac{9x^3}{3} + k \)
Simplify the expression:
\( R = 14x - 3x^2 + 3x^3 + k \)
Assuming that when no units are produced (x=0), there is no revenue (R=0), we can find the constant of integration (k):
\( 0 = 14(0) - 3(0)^2 + 3(0)^3 + k \)
\( \implies k = 0 \)
So, the total revenue function is:
\( R = 14x - 3x^2 + 3x^3 \)
The demand function (P), which represents the price per unit, is found by dividing the total revenue (R) by the number of units (x):
\( P = \frac{R}{x} \)
\( P = \frac{14x - 3x^2 + 3x^3}{x} \)
Divide each term by x:
\( P = 14 - 3x + 3x^2 \)
In simple words: We first found the total money earned by integrating the marginal revenue. Then, we figured out the demand function, which tells us the price for each item, by simply dividing the total money earned by the number of items sold.

🎯 Exam Tip: Remember the three key relationships: marginal revenue is the derivative of total revenue, total revenue is the integral of marginal revenue, and the demand function (price) is total revenue divided by quantity.