Samacheer Kalvi Class 11 Business Maths Solutions Chapter 6 Applications of Differentiation Exercise 6.1

Samacheer Kalvi 11th Business Maths Applications of Differentiation Ex 6.1 Text Book Back Questions and Answers

Question 1. If x tonnes of output at a total cost of \( C(x) = \frac{1}{10} x^3 – 4x^2 – 20x + 7 \), find the
(i) average cost
(ii) average variable cost
(iii) average fixed cost
(iv) marginal cost and
(v) marginal average cost.
Answer:
Given the total cost function: \( C(x) = \frac{1}{10} x^3 – 4x^2 – 20x + 7 \)
We know that total cost \( C(x) \) is made of variable cost \( f(x) \) and fixed cost \( k \).
So, \( C(x) = f(x) + k \)
From the given function, we can see that \( f(x) = \frac{1}{10} x^3 – 4x^2 – 20x \) and \( k = 7 \).

(i) **Average Cost (AC)**
Average Cost is calculated by dividing the total cost by the output \( x \).
\( AC = \frac{C(x)}{x} \)
\( AC = \frac{\frac{1}{10} x^3 – 4x^2 – 20x + 7}{x} \)
\( AC = \frac{1}{10} x^2 – 4x – 20 + \frac{7}{x} \)

(ii) **Average Variable Cost (AVC)**
Average Variable Cost is calculated by dividing the variable cost by the output \( x \).
\( AVC = \frac{f(x)}{x} \)
\( AVC = \frac{\frac{1}{10} x^3 – 4x^2 – 20x}{x} \)
\( AVC = \frac{1}{10} x^2 – 4x – 20 \)

(iii) **Average Fixed Cost (AFC)**
Average Fixed Cost is calculated by dividing the fixed cost by the output \( x \).
\( AFC = \frac{k}{x} \)
\( AFC = \frac{7}{x} \)

(iv) **Marginal Cost (MC)**
Marginal Cost is the derivative of the total cost function with respect to \( x \).
\( MC = \frac{dC}{dx} \)
\( MC = \frac{d}{dx} \left( \frac{1}{10} x^3 – 4x^2 – 20x + 7 \right) \)
\( MC = \frac{1}{10} (3x^{3-1}) – 4(2x^{2-1}) – 20(1) + 0 \)
\( MC = \frac{3}{10} x^2 – 8x – 20 \)

(v) **Marginal Average Cost (MAC)**
Marginal Average Cost is the derivative of the Average Cost function with respect to \( x \).
\( MAC = \frac{d}{dx} (AC) \)
\( MAC = \frac{d}{dx} \left( \frac{1}{10} x^2 – 4x – 20 + \frac{7}{x} \right) \)
\( MAC = \frac{1}{10} (2x) – 4(1) – 0 – \frac{7}{x^2} \)
\( MAC = \frac{1}{5} x – 4 – \frac{7}{x^2} \)
In simple words: This problem shows how different types of costs change when a company produces more goods. Average cost is the total cost per item, while marginal cost is the cost to make one extra item. Knowing these helps a business decide how much to produce.

🎯 Exam Tip: Remember to differentiate each term in the cost function carefully. Fixed costs become zero when differentiated, as they don't change with output.

Question 2. The total cost of x units of output of a firm is given by \( C = \frac{2}{3} x + \frac{35}{2} \). Find the
(i) cost when output is 4 units
(ii) average cost when output is 10 units
(iii) marginal cost when output is 3 units
Answer:
Given the total cost function: \( C(x) = \frac{2}{3} x + \frac{35}{2} \)

(i) **Cost when output is 4 units**
To find the cost when \( x = 4 \), substitute 4 into the cost function:
\( C(4) = \frac{2}{3} (4) + \frac{35}{2} \)
\( C(4) = \frac{8}{3} + \frac{35}{2} \)
To add these fractions, find a common denominator, which is 6.
\( C(4) = \frac{8 \times 2}{3 \times 2} + \frac{35 \times 3}{2 \times 3} \)
\( C(4) = \frac{16}{6} + \frac{105}{6} \)
\( C(4) = \frac{16 + 105}{6} \)
\( C(4) = \frac{121}{6} \)

(ii) **Average cost when output is 10 units**
First, find the average cost function, which is total cost divided by output \( x \).
\( AC(x) = \frac{C(x)}{x} \)
\( AC(x) = \frac{\frac{2}{3} x + \frac{35}{2}}{x} \)
\( AC(x) = \frac{2}{3} + \frac{35}{2x} \)
Now, substitute \( x = 10 \) into the average cost function:
\( AC(10) = \frac{2}{3} + \frac{35}{2(10)} \)
\( AC(10) = \frac{2}{3} + \frac{35}{20} \)
To add these fractions, find a common denominator, which is 60. Simplify \( \frac{35}{20} \) to \( \frac{7}{4} \) first.
\( AC(10) = \frac{2}{3} + \frac{7}{4} \)
\( AC(10) = \frac{2 \times 4}{3 \times 4} + \frac{7 \times 3}{4 \times 3} \)
\( AC(10) = \frac{8}{12} + \frac{21}{12} \)
\( AC(10) = \frac{8 + 21}{12} \)
\( AC(10) = \frac{29}{12} \)

(iii) **Marginal cost when output is 3 units**
Marginal cost is the derivative of the total cost function with respect to \( x \).
\( MC(x) = \frac{dC}{dx} \)
\( MC(x) = \frac{d}{dx} \left( \frac{2}{3} x + \frac{35}{2} \right) \)
\( MC(x) = \frac{2}{3} (1) + 0 \)
\( MC(x) = \frac{2}{3} \)
Since the marginal cost is a constant, it will be \( \frac{2}{3} \) even when output is 3 units.
\( MC(3) = \frac{2}{3} \)
In simple words: We calculated how much it costs to make a certain number of items, how much each item costs on average, and how much it costs to make just one more item. These calculations are very useful for managing production and pricing in a business.

🎯 Exam Tip: When the total cost function is linear (like \( C = ax + b \)), the marginal cost is always equal to the variable cost per unit (a in this case) and remains constant, regardless of the output level.

Question 3. Revenue function 'R' and cost function 'C' are \( R = 14x – x^2 \) and \( C = x(x^2 – 2) \). Find the
(i) average cost
(ii) marginal cost
(iii) average revenue and
(iv) marginal revenue.
Answer:
Given revenue function: \( R(x) = 14x – x^2 \)
Given cost function: \( C(x) = x(x^2 – 2) = x^3 – 2x \)

(i) **Average Cost (AC)**
Average Cost is total cost divided by output \( x \).
\( AC = \frac{C(x)}{x} \)
\( AC = \frac{x^3 – 2x}{x} \)
\( AC = x^2 – 2 \)

(ii) **Marginal Cost (MC)**
Marginal Cost is the derivative of the total cost function with respect to \( x \).
\( MC = \frac{dC}{dx} \)
\( MC = \frac{d}{dx} (x^3 – 2x) \)
\( MC = 3x^2 – 2 \)

(iii) **Average Revenue (AR)**
Average Revenue is total revenue divided by output \( x \).
\( AR = \frac{R(x)}{x} \)
\( AR = \frac{14x – x^2}{x} \)
\( AR = 14 – x \)

(iv) **Marginal Revenue (MR)**
Marginal Revenue is the derivative of the total revenue function with respect to \( x \).
\( MR = \frac{dR}{dx} \)
\( MR = \frac{d}{dx} (14x – x^2) \)
\( MR = 14 – 2x \)
In simple words: This problem helps us understand how a company's costs and earnings change with the number of items it sells. Average values show per-item amounts, while marginal values show the change from selling one more item, which is key for making business decisions.

🎯 Exam Tip: Always expand the cost or revenue functions first if they are in factored form, like \( C = x(x^2 - 2) \), before taking derivatives for marginal functions.

Question 4. If the demand law is given by \( p = 10e^{-\frac{x}{2}} \) then find the elasticity of demand.
Answer:
Given the demand law: \( p = 10e^{-\frac{x}{2}} \)
The formula for elasticity of demand \( \eta_d \) is \( \eta_d = -\frac{p}{x} \cdot \frac{dx}{dp} \).
First, we need to find \( \frac{dp}{dx} \).
\( \frac{dp}{dx} = \frac{d}{dx} \left( 10e^{-\frac{x}{2}} \right) \)
\( \frac{dp}{dx} = 10 \cdot e^{-\frac{x}{2}} \cdot \frac{d}{dx} \left( -\frac{x}{2} \right) \)
\( \frac{dp}{dx} = 10 \cdot e^{-\frac{x}{2}} \cdot \left( -\frac{1}{2} \right) \)
\( \frac{dp}{dx} = -5e^{-\frac{x}{2}} \)
Now, we need \( \frac{dx}{dp} \), which is the reciprocal of \( \frac{dp}{dx} \).
\( \frac{dx}{dp} = \frac{1}{-5e^{-\frac{x}{2}}} \)
Substitute \( p \), \( x \), and \( \frac{dx}{dp} \) into the elasticity of demand formula:
\( \eta_d = -\frac{p}{x} \cdot \frac{dx}{dp} \)
\( \eta_d = -\frac{10e^{-\frac{x}{2}}}{x} \cdot \left( \frac{1}{-5e^{-\frac{x}{2}}} \right) \)
\( \eta_d = -\frac{10e^{-\frac{x}{2}}}{x} \cdot \left( -\frac{1}{5e^{-\frac{x}{2}}} \right) \)
\( \eta_d = \frac{10e^{-\frac{x}{2}}}{5xe^{-\frac{x}{2}}} \)
\( \eta_d = \frac{10}{5x} \)
\( \eta_d = \frac{2}{x} \)
In simple words: Elasticity of demand tells us how much the quantity demanded changes when the price changes. Here, we found that it depends on the quantity 'x' itself. This means that at different levels of demand, the product will respond differently to price changes.

🎯 Exam Tip: Be careful with the chain rule when differentiating exponential functions. Remember that \( \frac{d}{dx} e^{f(x)} = e^{f(x)} \cdot f'(x) \).

Question 5. Find the elasticity of demand in terms of x for the following demand laws and also find the value of x where elasticity is equals to unity.
(i) \( p = (a – bx)^2 \)
(ii) \( p = a - bx^2 \)
Answer:
(i) **Given demand law: \( p = (a – bx)^2 \)**
First, find \( \frac{dp}{dx} \).
\( \frac{dp}{dx} = \frac{d}{dx} (a – bx)^2 \)
\( \frac{dp}{dx} = 2(a – bx)^{2-1} \cdot \frac{d}{dx} (a – bx) \)
\( \frac{dp}{dx} = 2(a – bx) (0 – b) \)
\( \frac{dp}{dx} = -2b(a – bx) \)
Now, calculate the elasticity of demand \( \eta_d \). The formula is \( \eta_d = -\frac{p}{x} \cdot \frac{dx}{dp} \).
\( \eta_d = -\frac{p}{x} \cdot \frac{1}{\frac{dp}{dx}} \)
\( \eta_d = -\frac{(a – bx)^2}{x} \cdot \frac{1}{-2b(a – bx)} \)
\( \eta_d = \frac{(a – bx)^2}{x \cdot 2b(a – bx)} \)
\( \eta_d = \frac{a – bx}{2bx} \)

**Value of x when elasticity is unity (\( \eta_d = 1 \)):**
Set the elasticity expression equal to 1:
\( \frac{a – bx}{2bx} = 1 \)
\( a – bx = 2bx \)
\( a = 2bx + bx \)
\( a = 3bx \)
\( x = \frac{a}{3b} \)

(ii) **Given demand law: \( p = a - bx^2 \)**
First, find \( \frac{dp}{dx} \).
\( \frac{dp}{dx} = \frac{d}{dx} (a - bx^2) \)
\( \frac{dp}{dx} = 0 - b(2x) \)
\( \frac{dp}{dx} = -2bx \)
Now, calculate the elasticity of demand \( \eta_d \). The formula is \( \eta_d = -\frac{p}{x} \cdot \frac{dx}{dp} \).
\( \eta_d = -\frac{p}{x} \cdot \frac{1}{\frac{dp}{dx}} \)
\( \eta_d = -\frac{a - bx^2}{x} \cdot \frac{1}{-2bx} \)
\( \eta_d = \frac{a - bx^2}{2bx^2} \)

**Value of x when elasticity is unity (\( \eta_d = 1 \)):**
Set the elasticity expression equal to 1:
\( \frac{a - bx^2}{2bx^2} = 1 \)
\( a - bx^2 = 2bx^2 \)
\( a = 2bx^2 + bx^2 \)
\( a = 3bx^2 \)
\( x^2 = \frac{a}{3b} \)
\( x = \sqrt{\frac{a}{3b}} \)
In simple words: We calculated how sensitive the demand for a product is to price changes for two different demand formulas. When elasticity is 1, it means a percentage change in price causes the same percentage change in quantity demanded. This 'unity' point is important for businesses to maximize revenue.

🎯 Exam Tip: Remember to use the chain rule for derivatives like \( (a-bx)^2 \). Pay close attention to negative signs, as they are crucial in elasticity calculations.

Question 6. Find the elasticity of supply for the supply function \( x = 2p^2 + 5 \) when \( p = 3 \).
Answer:
Given the supply function: \( x = 2p^2 + 5 \)
The formula for elasticity of supply \( \eta_s \) is \( \eta_s = \frac{p}{x} \cdot \frac{dx}{dp} \).
First, find \( \frac{dx}{dp} \).
\( \frac{dx}{dp} = \frac{d}{dp} (2p^2 + 5) \)
\( \frac{dx}{dp} = 2(2p) + 0 \)
\( \frac{dx}{dp} = 4p \)
Now, substitute \( x \) and \( \frac{dx}{dp} \) into the elasticity of supply formula:
\( \eta_s = \frac{p}{x} \cdot \frac{dx}{dp} \)
\( \eta_s = \frac{p}{2p^2 + 5} \cdot (4p) \)
\( \eta_s = \frac{4p^2}{2p^2 + 5} \)
Now, find the elasticity of supply when \( p = 3 \).
\( \eta_s = \frac{4(3)^2}{2(3)^2 + 5} \)
\( \eta_s = \frac{4 \times 9}{2 \times 9 + 5} \)
\( \eta_s = \frac{36}{18 + 5} \)
\( \eta_s = \frac{36}{23} \)
In simple words: Elasticity of supply shows how much the quantity of goods a producer offers changes when the price changes. When the price is 3, the supply is not very elastic, meaning that a change in price will not lead to a huge change in the amount supplied.

🎯 Exam Tip: Remember that elasticity of supply \( \eta_s \) is generally positive, as producers usually supply more when prices increase. Demand elasticity, \( \eta_d \), is often negative but usually reported as its absolute value.

Question 7. The demand curve of a commodity is given by \( p = \frac{50-x}{5} \), find the marginal revenue for any output x and also find marginal revenue at \( x = 0 \) and \( x = 25 \)?
Answer:
Given the demand curve: \( p = \frac{50-x}{5} \)
First, find the total revenue function \( R(x) \). Revenue is price multiplied by quantity.
\( R(x) = p \cdot x \)
\( R(x) = \left( \frac{50-x}{5} \right) x \)
\( R(x) = \frac{50x - x^2}{5} \)
\( R(x) = \frac{1}{5} (50x - x^2) \)
Now, find the marginal revenue (MR) by taking the derivative of the total revenue function with respect to \( x \).
\( MR = \frac{dR}{dx} \)
\( MR = \frac{d}{dx} \left( \frac{1}{5} (50x - x^2) \right) \)
\( MR = \frac{1}{5} (50 - 2x) \)

**Marginal Revenue when \( x = 0 \):**
Substitute \( x = 0 \) into the MR function:
\( MR(0) = \frac{1}{5} (50 - 2(0)) \)
\( MR(0) = \frac{1}{5} (50) \)
\( MR(0) = 10 \)

**Marginal Revenue when \( x = 25 \):**
Substitute \( x = 25 \) into the MR function:
\( MR(25) = \frac{1}{5} (50 - 2(25)) \)
\( MR(25) = \frac{1}{5} (50 - 50) \)
\( MR(25) = \frac{1}{5} (0) \)
\( MR(25) = 0 \)
In simple words: Marginal revenue is the extra money a company gets from selling one more unit. We found a formula for it, and then saw that selling the first unit brings in Rs. 10, but selling a unit when 25 are already being sold brings in no extra money. This means the company might not want to produce beyond 25 units.

🎯 Exam Tip: Remember to always first find the total revenue function \( R = p \cdot x \) before differentiating to find marginal revenue. Sometimes \( p \) is given as a function of \( x \).

Question 8. The supply function of certain goods is given by \( x = a\sqrt{p-b} \) where p is unit price, a and b are constants with \( p > b \). Find elasticity of supply at \( p = 2b \).
Answer:
Given the supply function: \( x = a\sqrt{p-b} = a(p-b)^{1/2} \)
The formula for elasticity of supply \( \eta_s \) is \( \eta_s = \frac{p}{x} \cdot \frac{dx}{dp} \).
First, find \( \frac{dx}{dp} \).
\( \frac{dx}{dp} = \frac{d}{dp} (a(p-b)^{1/2}) \)
\( \frac{dx}{dp} = a \cdot \frac{1}{2} (p-b)^{1/2 - 1} \cdot \frac{d}{dp} (p-b) \)
\( \frac{dx}{dp} = a \cdot \frac{1}{2} (p-b)^{-1/2} \cdot (1) \)
\( \frac{dx}{dp} = \frac{a}{2\sqrt{p-b}} \)
Now, substitute \( x \) and \( \frac{dx}{dp} \) into the elasticity of supply formula:
\( \eta_s = \frac{p}{x} \cdot \frac{dx}{dp} \)
\( \eta_s = \frac{p}{a\sqrt{p-b}} \cdot \frac{a}{2\sqrt{p-b}} \)
\( \eta_s = \frac{p \cdot a}{a\sqrt{p-b} \cdot 2\sqrt{p-b}} \)
\( \eta_s = \frac{p}{2(p-b)} \)
Now, find the elasticity of supply when \( p = 2b \).
\( \eta_s = \frac{2b}{2(2b - b)} \)
\( \eta_s = \frac{2b}{2(b)} \)
\( \eta_s = \frac{2b}{2b} \)
\( \eta_s = 1 \)
In simple words: We found how supply reacts to price changes using a specific formula. When the price reaches twice the value of 'b', the supply elasticity is exactly 1. This means that a 1% change in price will cause a 1% change in the quantity supplied.

🎯 Exam Tip: When differentiating square root functions, remember to convert them to fractional exponents, e.g., \( \sqrt{u} = u^{1/2} \), then apply the power rule and chain rule.

Question 9. Show that \( MR = p \left[1-\frac{1}{\eta_d}\right] \) for the demand function \( p = 400 – 2x – 3x^2 \) where p is unit price and x is quantity demand.
Answer:
Given the demand function: \( p = 400 – 2x – 3x^2 \)
First, find the total revenue function \( R(x) \). Revenue is price multiplied by quantity.
\( R(x) = p \cdot x \)
\( R(x) = (400 – 2x – 3x^2)x \)
\( R(x) = 400x – 2x^2 – 3x^3 \)
Now, find the marginal revenue (MR) by taking the derivative of the total revenue function with respect to \( x \).
\( MR = \frac{dR}{dx} \)
\( MR = \frac{d}{dx} (400x – 2x^2 – 3x^3) \)
\( MR = 400 – 4x – 9x^2 \)

Next, find the elasticity of demand \( \eta_d \). First, find \( \frac{dp}{dx} \).
\( \frac{dp}{dx} = \frac{d}{dx} (400 – 2x – 3x^2) \)
\( \frac{dp}{dx} = 0 – 2 – 6x \)
\( \frac{dp}{dx} = -2 – 6x \)
The formula for elasticity of demand \( \eta_d \) is \( \eta_d = -\frac{p}{x} \cdot \frac{dx}{dp} \).
\( \eta_d = -\frac{p}{x} \cdot \frac{1}{\frac{dp}{dx}} \)
\( \eta_d = -\frac{400 – 2x – 3x^2}{x} \cdot \frac{1}{-2 – 6x} \)
\( \eta_d = \frac{400 – 2x – 3x^2}{x(2 + 6x)} \)
\( \eta_d = \frac{400 – 2x – 3x^2}{2x(1 + 3x)} \)

Now, substitute \( \eta_d \) into the expression \( p \left[1-\frac{1}{\eta_d}\right] \).
\( p \left[1-\frac{1}{\eta_d}\right] = (400 – 2x – 3x^2) \left[1 - \frac{2x(1 + 3x)}{400 – 2x – 3x^2}\right] \)
\( = (400 – 2x – 3x^2) \left[\frac{(400 – 2x – 3x^2) - 2x(1 + 3x)}{400 – 2x – 3x^2}\right] \)
\( = (400 – 2x – 3x^2) - 2x(1 + 3x) \)
\( = 400 – 2x – 3x^2 - 2x - 6x^2 \)
\( = 400 – 4x – 9x^2 \)
This result is equal to the MR we calculated earlier.
Therefore, \( MR = p \left[1-\frac{1}{\eta_d}\right] \) is shown to be true.
In simple words: We've proven a math rule that connects marginal revenue to price and demand elasticity. This formula is very helpful for businesses to understand how changes in price affect their total earnings, especially when demand is elastic or inelastic.

🎯 Exam Tip: When asked to "show that" a formula is true, you must derive both sides independently or manipulate one side to match the other. Always clearly state your steps.

Question 10. For the demand function \( p = 550 – 3x – 6x^2 \) where x is quantity demand and p is unit price. Show that \( MR = p \left[1-\frac{1}{\eta_d}\right] \).
Answer:
Given the demand function: \( p = 550 – 3x – 6x^2 \)
First, find the total revenue function \( R(x) \). Revenue is price multiplied by quantity.
\( R(x) = p \cdot x \)
\( R(x) = (550 – 3x – 6x^2)x \)
\( R(x) = 550x – 3x^2 – 6x^3 \)
Now, find the marginal revenue (MR) by taking the derivative of the total revenue function with respect to \( x \).
\( MR = \frac{dR}{dx} \)
\( MR = \frac{d}{dx} (550x – 3x^2 – 6x^3) \)
\( MR = 550 – 6x – 18x^2 \)

Next, find the elasticity of demand \( \eta_d \). First, find \( \frac{dp}{dx} \).
\( \frac{dp}{dx} = \frac{d}{dx} (550 – 3x – 6x^2) \)
\( \frac{dp}{dx} = 0 – 3 – 12x \)
\( \frac{dp}{dx} = -3 – 12x \)
The formula for elasticity of demand \( \eta_d \) is \( \eta_d = -\frac{p}{x} \cdot \frac{dx}{dp} \).
\( \eta_d = -\frac{p}{x} \cdot \frac{1}{\frac{dp}{dx}} \)
\( \eta_d = -\frac{550 – 3x – 6x^2}{x} \cdot \frac{1}{-3 – 12x} \)
\( \eta_d = \frac{550 – 3x – 6x^2}{x(3 + 12x)} \)
\( \eta_d = \frac{550 – 3x – 6x^2}{3x(1 + 4x)} \)

Now, substitute \( \eta_d \) into the expression \( p \left[1-\frac{1}{\eta_d}\right] \).
\( p \left[1-\frac{1}{\eta_d}\right] = (550 – 3x – 6x^2) \left[1 - \frac{3x(1 + 4x)}{550 – 3x – 6x^2}\right] \)
\( = (550 – 3x – 6x^2) \left[\frac{(550 – 3x – 6x^2) - 3x(1 + 4x)}{550 – 3x – 6x^2}\right] \)
\( = (550 – 3x – 6x^2) - 3x(1 + 4x) \)
\( = 550 – 3x – 6x^2 - 3x - 12x^2 \)
\( = 550 – 6x – 18x^2 \)
This result is equal to the MR we calculated earlier.
Therefore, \( MR = p \left[1-\frac{1}{\eta_d}\right] \) is shown to be true.
In simple words: This problem confirms an important relationship in economics: how marginal revenue is linked to price and how sensitive demand is to price changes. By using the given demand function, we derived the formula, which helps businesses understand their pricing strategies.

🎯 Exam Tip: When simplifying fractions with polynomial terms, look for common factors in the numerator and denominator to reduce complexity before performing further calculations.

Question 11. For the demand function \( x = \frac{25}{p^4} \), \( 1 \le p \le 5 \), determine the elasticity of demand.
Answer:
Given the demand function: \( x = \frac{25}{p^4} = 25p^{-4} \)
The formula for elasticity of demand \( \eta_d \) is \( \eta_d = -\frac{p}{x} \cdot \frac{dx}{dp} \).
First, find \( \frac{dx}{dp} \).
\( \frac{dx}{dp} = \frac{d}{dp} (25p^{-4}) \)
\( \frac{dx}{dp} = 25 \cdot (-4)p^{-4-1} \)
\( \frac{dx}{dp} = -100p^{-5} \)
\( \frac{dx}{dp} = -\frac{100}{p^5} \)
Now, substitute \( x \) and \( \frac{dx}{dp} \) into the elasticity of demand formula:
\( \eta_d = -\frac{p}{x} \cdot \frac{dx}{dp} \)
\( \eta_d = -\frac{p}{\frac{25}{p^4}} \cdot \left( -\frac{100}{p^5} \right) \)
\( \eta_d = -\frac{p \cdot p^4}{25} \cdot \left( -\frac{100}{p^5} \right) \)
\( \eta_d = -\frac{p^5}{25} \cdot \left( -\frac{100}{p^5} \right) \)
\( \eta_d = \frac{100p^5}{25p^5} \)
\( \eta_d = 4 \)
In simple words: We calculated how much demand changes when the price changes for a specific demand function. In this case, the elasticity of demand is 4, which means that for every 1% change in price, the quantity demanded will change by 4%. This indicates a very elastic demand.

🎯 Exam Tip: When dealing with elasticity calculations, expressing functions with negative exponents (e.g., \( \frac{1}{p^4} = p^{-4} \)) can simplify differentiation using the power rule.

Question 12. The demand function of a commodity is \( p = 200 – \frac{x}{100} \) and its cost is \( C = 40x + 120 \) where p is a unit price in rupees and x is the number of units produced and sold. Determine
(i) profit function
(ii) average profit at an output of 10 units
(iii) marginal profit at an output of 10 units and
(iv) marginal average profit at an output of 10 units.
Answer:
Given demand function: \( p = 200 – \frac{x}{100} \)
Given cost function: \( C(x) = 40x + 120 \)

First, find the total revenue function \( R(x) \).
\( R(x) = p \cdot x \)
\( R(x) = \left( 200 – \frac{x}{100} \right) x \)
\( R(x) = 200x – \frac{x^2}{100} \)

(i) **Profit function \( P(x) \)**
Profit is total revenue minus total cost.
\( P(x) = R(x) – C(x) \)
\( P(x) = \left( 200x – \frac{x^2}{100} \right) – (40x + 120) \)
\( P(x) = 200x – \frac{x^2}{100} – 40x – 120 \)
\( P(x) = 160x – \frac{x^2}{100} – 120 \)

(ii) **Average profit (AP) at an output of 10 units**
Average profit is total profit divided by output \( x \).
\( AP(x) = \frac{P(x)}{x} \)
\( AP(x) = \frac{160x – \frac{x^2}{100} – 120}{x} \)
\( AP(x) = 160 – \frac{x}{100} – \frac{120}{x} \)
Now, substitute \( x = 10 \) into the average profit function:
\( AP(10) = 160 – \frac{10}{100} – \frac{120}{10} \)
\( AP(10) = 160 – 0.1 – 12 \)
\( AP(10) = 148 – 0.1 \)
\( AP(10) = \text{Rs. } 147.9 \)

(iii) **Marginal profit (MP) at an output of 10 units**
Marginal profit is the derivative of the total profit function with respect to \( x \).
\( MP(x) = \frac{dP}{dx} \)
\( MP(x) = \frac{d}{dx} \left( 160x – \frac{x^2}{100} – 120 \right) \)
\( MP(x) = 160 – \frac{2x}{100} – 0 \)
\( MP(x) = 160 – \frac{x}{50} \)
Now, substitute \( x = 10 \) into the marginal profit function:
\( MP(10) = 160 – \frac{10}{50} \)
\( MP(10) = 160 – \frac{1}{5} \)
\( MP(10) = 160 – 0.2 \)
\( MP(10) = \text{Rs. } 159.8 \)

(iv) **Marginal average profit (MAP) at an output of 10 units**
Marginal average profit is the derivative of the average profit function with respect to \( x \).
\( MAP(x) = \frac{d}{dx} (AP(x)) \)
\( MAP(x) = \frac{d}{dx} \left( 160 – \frac{x}{100} – \frac{120}{x} \right) \)
\( MAP(x) = \frac{d}{dx} \left( 160 – \frac{1}{100}x – 120x^{-1} \right) \)
\( MAP(x) = 0 – \frac{1}{100} – 120(-1)x^{-1-1} \)
\( MAP(x) = -\frac{1}{100} + 120x^{-2} \)
\( MAP(x) = -\frac{1}{100} + \frac{120}{x^2} \)
Now, substitute \( x = 10 \) into the marginal average profit function:
\( MAP(10) = -\frac{1}{100} + \frac{120}{(10)^2} \)
\( MAP(10) = -\frac{1}{100} + \frac{120}{100} \)
\( MAP(10) = \frac{-1 + 120}{100} \)
\( MAP(10) = \frac{119}{100} \)
\( MAP(10) = \text{Rs. } 1.19 \)
In simple words: This problem shows how to calculate profit, average profit, marginal profit, and marginal average profit for a company. These are important tools that help a business understand its overall financial health and how each extra unit produced affects its earnings and efficiency.

🎯 Exam Tip: Distinguish between total, average, and marginal functions. Total refers to the overall amount, average is per unit, and marginal is the change from one additional unit. Derivatives are used to find marginal functions.

Question 13. Find the values of x, when the marginal function of \( y = x^3 + 10x^2 - 48x + 8 \) is twice the x.
Answer: The given function is \( y = x^3 + 10x^2 - 48x + 8 \).
First, we find the marginal function by taking the derivative of y with respect to x.
Marginal function \( \frac{d y}{d x} = \frac{d}{d x}(x^3 + 10x^2 - 48x + 8) \)
\( = 3x^{3-1} + 10(2)x^{2-1} - 48(1) + 0 \)
\( = 3x^2 + 20x - 48 \)
We are told that the marginal function is twice the value of x.
So, we set the marginal function equal to \( 2x \):
\( 3x^2 + 20x - 48 = 2x \)
Now, we rearrange the equation to form a standard quadratic equation:
\( 3x^2 + 20x - 2x - 48 = 0 \)
\( 3x^2 + 18x - 48 = 0 \)
To simplify, we divide the entire equation by 3:
\( \frac{3x^2}{3} + \frac{18x}{3} - \frac{48}{3} = \frac{0}{3} \)
\( x^2 + 6x - 16 = 0 \)
Next, we factor the quadratic equation. We need two numbers that multiply to -16 and add to 6. These numbers are 8 and -2.
\( (x + 8)(x - 2) = 0 \)
This gives us two possible values for x:
\( x + 8 = 0 \implies x = -8 \)
or
\( x - 2 = 0 \implies x = 2 \)
So, the values of x are -8 and 2. Analyzing the marginal function often helps firms understand how their costs or revenues change with each extra unit produced.
In simple words: First, we find the rate of change of the function. Then, we set this rate equal to two times x and solve the equation. The two numbers that x can be are -8 and 2.

🎯 Exam Tip: Remember to simplify the quadratic equation by dividing by a common factor before factoring, as it makes the process easier and reduces errors.

Question 14. The total cost function y for x units is given by \( y = 3x\left(\frac{x+7}{x+5}\right) + 5 \). Show that the marginal cost decreases continuously as the output increases.
Answer: The total cost function is given as \( y = 3x\left(\frac{x+7}{x+5}\right) + 5 \).
We can rewrite the fraction \( \frac{x+7}{x+5} \) by adding and subtracting 5 in the numerator:
\( \frac{x+7}{x+5} = \frac{(x+5)+2}{x+5} = 1 + \frac{2}{x+5} \)
Substitute this back into the cost function:
\( y = 3x\left(1 + \frac{2}{x+5}\right) + 5 \)
\( y = 3x + \frac{6x}{x+5} + 5 \)
To show that the marginal cost decreases continuously as output increases, we need to find the marginal cost (which is the first derivative of y with respect to x, \( \frac{dy}{dx} \)) and then examine its rate of change (the second derivative, \( \frac{d^2y}{dx^2} \)). If the second derivative is negative, it means the marginal cost is decreasing.
First, let's find the marginal cost \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{d}{dx}\left(3x + \frac{6x}{x+5} + 5\right) \)
\( \frac{dy}{dx} = \frac{d}{dx}(3x) + \frac{d}{dx}\left(\frac{6x}{x+5}\right) + \frac{d}{dx}(5) \)
\( = 3 + 6 \frac{d}{dx}\left(\frac{x}{x+5}\right) + 0 \)
We use the quotient rule for \( \frac{d}{dx}\left(\frac{x}{x+5}\right) \): \( \frac{u'v - uv'}{v^2} \) where \( u = x, u' = 1, v = x+5, v' = 1 \).
So, \( \frac{d}{dx}\left(\frac{x}{x+5}\right) = \frac{(1)(x+5) - x(1)}{(x+5)^2} = \frac{x+5-x}{(x+5)^2} = \frac{5}{(x+5)^2} \)
Therefore, the marginal cost is:
\( \frac{dy}{dx} = 3 + 6 \left(\frac{5}{(x+5)^2}\right) \)
\( = 3 + \frac{30}{(x+5)^2} \)
Now, we need to find the second derivative to see if the marginal cost is decreasing.
\( \frac{d^2y}{dx^2} = \frac{d}{dx}\left(3 + \frac{30}{(x+5)^2}\right) \)
\( = \frac{d}{dx}(3) + \frac{d}{dx}\left(30(x+5)^{-2}\right) \)
\( = 0 + 30(-2)(x+5)^{-3}(1) \)
\( = -60(x+5)^{-3} \)
\( = \frac{-60}{(x+5)^3} \)
For any positive output \( x \), \( (x+5)^3 \) will be a positive number. Since the numerator is -60, the second derivative \( \frac{d^2y}{dx^2} \) will always be negative.
Since the second derivative of the total cost function is negative, it means the marginal cost is decreasing as the output (x) increases. This analysis helps businesses decide the optimal production levels.
In simple words: We first find how the cost changes for each extra unit made, which is the marginal cost. Then, we check if this marginal cost is going down as we make more units. We do this by finding the second derivative, and because it is negative, it means the marginal cost is indeed decreasing.

🎯 Exam Tip: To show that a function is continuously decreasing, calculate its second derivative and demonstrate that it is consistently negative for the relevant domain of the variable.

Question 15. Find the price elasticity of demand for the demand function \( x = 10 – p \) where x is the demand p is the price. Examine whether the demand is elastic, inelastic, or unit elastic at p = 6.
Answer: The demand function is given as \( x = 10 - p \).
To find the price elasticity of demand (\( \eta_d \)), we first need to find \( \frac{dx}{dp} \).
\( \frac{dx}{dp} = \frac{d}{dp}(10 - p) \)
\( = 0 - 1 \)
\( = -1 \)
The formula for price elasticity of demand is: \( \eta_d = -\frac{p}{x} \cdot \frac{dx}{dp} \)
Substitute the known values:
\( \eta_d = -\frac{p}{(10 - p)} \cdot (-1) \)
\( \eta_d = \frac{p}{10 - p} \)
Now, we need to examine the elasticity at \( p = 6 \).
Substitute \( p = 6 \) into the elasticity formula:
\( \eta_d = \frac{6}{10 - 6} \)
\( = \frac{6}{4} \)
\( = 1.5 \)
Since \( |\eta_d| = 1.5 > 1 \), the demand is elastic at \( p = 6 \). This means that a small change in price will lead to a larger percentage change in the quantity demanded.
In simple words: We find how much demand changes when the price changes. Then, we use a formula to calculate a number called elasticity. If this number is greater than 1, like 1.5, it means that at a price of 6, the demand is very sensitive to price changes.

🎯 Exam Tip: Remember that demand is elastic if \( |\eta_d| > 1 \), inelastic if \( |\eta_d| < 1 \), and unit elastic if \( |\eta_d| = 1 \).

Question 16. Find the equilibrium price and equilibrium quantity for the following functions. Demand: \( x = 100 – 2p \) and supply: \( x = 3p-50 \).
Answer: The given demand function is \( x = 100 - 2p \).
The given supply function is \( x = 3p - 50 \).
At equilibrium, the quantity demanded equals the quantity supplied. This point is where the market is stable.
So, we set the demand function equal to the supply function:
\( 100 - 2p = 3p - 50 \)
Now, we solve for \( p \) (price). First, move all terms with \( p \) to one side and constants to the other:
\( -2p - 3p = -50 - 100 \)
\( -5p = -150 \)
Divide both sides by -5 to find \( p \):
\( p = \frac{-150}{-5} \)
\( p = 30 \)
So, the equilibrium price \( P_E \) is Rs. 30.
Next, we find the equilibrium quantity \( X_E \) by substituting the equilibrium price (\( p = 30 \)) into either the demand or supply function. Let's use the supply function:
\( x = 3p - 50 \)
\( x = 3(30) - 50 \)
\( x = 90 - 50 \)
\( x = 40 \)
So, the equilibrium quantity \( X_E \) is 40 units.
In simple words: We are given how much people want to buy (demand) and how much is available to sell (supply) at different prices. To find the balance point, we make demand and supply equal. We solve for the price first, which turns out to be Rs. 30. Then, we use this price to find how many items are bought and sold, which is 40 units.

🎯 Exam Tip: Always state the equilibrium price and quantity clearly with their respective units (e.g., Rs. for price, units for quantity).

Question 17. The demand and cost functions of a firm are \( x = 6000 – 30p \) and \( C = 72000 + 60x \) respectively. Find the level of output and price at which the profit is maximum.
Answer: We know that profit is at its maximum when the marginal revenue (MR) equals the marginal cost (MC). This is a key principle in economics for profit maximization.
Given the demand function: \( x = 6000 - 30p \)
We need to express price \( p \) in terms of \( x \):
\( 30p = 6000 - x \)
\( p = \frac{6000 - x}{30} \)
\( p = \frac{6000}{30} - \frac{x}{30} \)
\( p = 200 - \frac{x}{30} \) ... (1)
Next, we find the Revenue function \( R \). Revenue is Price times Quantity \( (R = px) \):
\( R = \left(200 - \frac{x}{30}\right)x \)
\( R = 200x - \frac{x^2}{30} \)
Now, we calculate the Marginal Revenue (MR) by taking the derivative of the Revenue function with respect to \( x \):
\( MR = \frac{dR}{dx} = \frac{d}{dx}\left(200x - \frac{x^2}{30}\right) \)
\( MR = 200(1) - \frac{1}{30}(2x) \)
\( MR = 200 - \frac{2x}{30} \)
\( MR = 200 - \frac{x}{15} \)
The Cost function is given as: \( C = 72000 + 60x \)
We calculate the Marginal Cost (MC) by taking the derivative of the Cost function with respect to \( x \):
\( MC = \frac{dC}{dx} = \frac{d}{dx}(72000 + 60x) \)
\( MC = 0 + 60(1) \)
\( MC = 60 \)
To find the level of output for maximum profit, we set Marginal Revenue equal to Marginal Cost:
\( MR = MC \)
\( 200 - \frac{x}{15} = 60 \)
Now, solve for \( x \):
\( 200 - 60 = \frac{x}{15} \)
\( 140 = \frac{x}{15} \)
\( x = 140 \times 15 \)
\( x = 2100 \)
So, the level of output for maximum profit is 2100 units.
Finally, to find the price at this output level, substitute \( x = 2100 \) into equation (1):
\( p = 200 - \frac{x}{30} \)
\( p = 200 - \frac{2100}{30} \)
\( p = 200 - 70 \)
\( p = 130 \)
Thus, the price for maximum profit is Rs. 130. By operating at the point where marginal revenue equals marginal cost, businesses can achieve the highest possible profit margin.
In simple words: To make the most profit, a company needs to find the point where the money earned from selling one extra item (marginal revenue) is equal to the cost of making that extra item (marginal cost). By setting these two equal, we found that the company should produce 2100 units and sell them at Rs. 130 each to get the highest profit.

🎯 Exam Tip: Always remember that profit maximization occurs when marginal revenue equals marginal cost. Make sure to clearly derive both MR and MC from the given functions.

Question 18. The cost function of a firm is \( C = x^3 – 12x^2 + 48x \). Find the level of output \( (x > 0) \) at which average cost is minimum.
Answer: The cost function of a firm is given by \( C = x^3 - 12x^2 + 48x \).
Average cost (AC) is minimum when Average Cost equals Marginal Cost (MC). This is an important concept for understanding efficiency.
First, let's find the Average Cost (AC). Average cost is total cost divided by the quantity \( x \):
\( AC = \frac{C}{x} = \frac{x^3 - 12x^2 + 48x}{x} \)
For \( x > 0 \), we can divide each term by \( x \):
\( AC = x^2 - 12x + 48 \)
Next, let's find the Marginal Cost (MC). Marginal cost is the derivative of the total cost function with respect to \( x \):
\( MC = \frac{dC}{dx} = \frac{d}{dx}(x^3 - 12x^2 + 48x) \)
\( MC = 3x^{3-1} - 12(2)x^{2-1} + 48(1) \)
\( MC = 3x^2 - 24x + 48 \)
To find the level of output where average cost is minimum, we set AC equal to MC:
\( AC = MC \)
\( x^2 - 12x + 48 = 3x^2 - 24x + 48 \)
Now, we solve for \( x \). Move all terms to one side of the equation:
\( x^2 - 3x^2 - 12x + 24x + 48 - 48 = 0 \)
\( -2x^2 + 12x = 0 \)
To simplify, divide the entire equation by -2:
\( \frac{-2x^2}{-2} + \frac{12x}{-2} = 0 \)
\( x^2 - 6x = 0 \)
Factor out \( x \):
\( x(x - 6) = 0 \)
This gives us two possible values for \( x \):
\( x = 0 \) or \( x - 6 = 0 \implies x = 6 \)
The question states that \( x > 0 \), so we disregard \( x = 0 \).
Therefore, the level of output at which average cost is minimum is \( x = 6 \) units. Producing at this level helps a firm operate efficiently.
In simple words: To find when the average cost of making each item is the lowest, we need to find the point where the average cost is equal to the cost of making just one more item (marginal cost). We set these two equal, solve the equation, and find that making 6 units gives the lowest average cost per item.

🎯 Exam Tip: Always remember that the average cost is minimized at the output level where the average cost curve intersects the marginal cost curve. Do not forget to check any domain restrictions (like \( x > 0 \)).