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

Question 1. Find (a) the averages and (b) the marginal cost given that for x units of commodity.
(i) \( C(x) = \frac { 1 }{ 3 }x³ + x^2 – 8x + 5 \)
(ii) \( C(x) = \frac{x^3}{3} + 3x^2 – 16x + 2 \)
(iii) \( C(x) = \frac{x^3}{3} + 3x^2 – 7x + 16 \)
Answer:
(i) Given cost function: \( C(x) = \frac{1}{3}x^3 + x^2 – 8x + 5 \)
(a) Marginal Cost Function (MC): We find this by taking the derivative of \( C(x) \) with respect to \( x \). The derivative of \( x^n \) is \( nx^{n-1} \).
\( MC = \frac{d}{dx} C(x) \)
\( MC = \frac{d}{dx} \left( \frac{1}{3}x^3 + x^2 – 8x + 5 \right) \)
\( MC = \frac{1}{3} (3x^2) + (2x) – 8 \)
\( MC = x^2 + 2x – 8 \)
(b) Average Cost Function (AC): We find this by dividing the total cost \( C(x) \) by the number of units \( x \).
\( AC = \frac{C(x)}{x} \)
\( AC = \frac{\frac{1}{3}x^3 + x^2 – 8x + 5}{x} \)
\( AC = \frac{1}{3}x^2 + x – 8 + \frac{5}{x} \)
(c) Slope of Average Cost Function: This is the derivative of the Average Cost Function with respect to \( x \).
\( \frac{d}{dx}(AC) = \frac{d}{dx} \left( \frac{1}{3}x^2 + x – 8 + \frac{5}{x} \right) \)
\( \frac{d}{dx}(AC) = \frac{1}{3}(2x) + 1 – 5x^{-2} \)
\( \frac{d}{dx}(AC) = \frac{2x}{3} + 1 – \frac{5}{x^2} \)

(ii) Given total cost function: \( C(x) = \frac{x^3}{3} + 3x^2 – 16x + 2 \)
(a) Marginal Cost Function (MC): We find this by taking the derivative of \( C(x) \) with respect to \( x \).
\( MC = \frac{d}{dx} C(x) \)
\( MC = \frac{d}{dx} \left( \frac{x^3}{3} + 3x^2 – 16x + 2 \right) \)
\( MC = \frac{3x^2}{3} + 3(2x) – 16 \)
\( MC = x^2 + 6x – 16 \)
(b) Average Cost Function (AC): We find this by dividing the total cost \( C(x) \) by the number of units \( x \).
\( AC = \frac{C(x)}{x} \)
\( AC = \frac{\frac{x^3}{3} + 3x^2 – 16x + 2}{x} \)
\( AC = \frac{x^2}{3} + 3x – 16 + \frac{2}{x} \)

(iii) Given total cost function: \( C(x) = \frac{x^3}{3} + 3x^2 – 7x + 16 \)
(a) Marginal Cost Function (MC): We find this by taking the derivative of \( C(x) \) with respect to \( x \).
\( MC = \frac{d}{dx} C(x) \)
\( MC = \frac{d}{dx} \left( \frac{x^3}{3} + 3x^2 – 7x + 16 \right) \)
\( MC = \frac{3x^2}{3} + 3(2x) – 7 \)
\( MC = x^2 + 6x – 7 \)
(b) Average Cost Function (AC): We find this by dividing the total cost \( C(x) \) by the number of units \( x \).
\( AC = \frac{C(x)}{x} \)
\( AC = \frac{\frac{x^3}{3} + 3x^2 – 7x + 16}{x} \)
\( AC = \frac{x^2}{3} + 3x – 7 + \frac{16}{x} \)
In simple words: To find the marginal cost, you take the derivative of the total cost function. This tells you the extra cost of making one more item. To find the average cost, you divide the total cost by the number of items made.

🎯 Exam Tip: Remember that marginal cost is always the first derivative of the total cost function, while average cost is the total cost divided by the quantity. Make sure to clearly state which part (a), (b), or (c) you are answering for each sub-question.

Question 2.
(i) If the total cost function for a manufacturer is given by \( C = \frac{5 x^2}{\sqrt{x^2+3}}+5000 \), find the marginal cost function.
(ii) The cost of producing x units of a product is given by \( C = 7x + 130 \), show that the marginal cost is always constant.
Answer:
(i) Given total cost function: \( C(x) = \frac{5 x^2}{\sqrt{x^2+3}}+5000 \)
To find the marginal cost function (MC), we need to calculate the derivative of \( C(x) \) with respect to \( x \).
\( MC = \frac{dC}{dx} = \frac{d}{dx} \left( \frac{5x^2}{\sqrt{x^2+3}} + 5000 \right) \)
We can separate the derivative of the sum:
\( MC = \frac{d}{dx} \left( \frac{5x^2}{\sqrt{x^2+3}} \right) + \frac{d}{dx} (5000) \)
The derivative of a constant (5000) is 0.
For the first term, we use the quotient rule, \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \).
Let \( u = 5x^2 \) and \( v = \sqrt{x^2+3} = (x^2+3)^{1/2} \).
Then \( u' = 10x \).
And \( v' = \frac{1}{2}(x^2+3)^{-1/2} (2x) = x(x^2+3)^{-1/2} = \frac{x}{\sqrt{x^2+3}} \).
Now substitute these into the quotient rule:
\( \frac{d}{dx} \left( \frac{5x^2}{\sqrt{x^2+3}} \right) = \frac{(10x)\sqrt{x^2+3} - (5x^2)\left(\frac{x}{\sqrt{x^2+3}}\right)}{(\sqrt{x^2+3})^2} \)
\( \frac{d}{dx} \left( \frac{5x^2}{\sqrt{x^2+3}} \right) = \frac{10x(x^2+3) - 5x^3}{(x^2+3)\sqrt{x^2+3}} \)
\( \frac{d}{dx} \left( \frac{5x^2}{\sqrt{x^2+3}} \right) = \frac{10x^3+30x - 5x^3}{(x^2+3)^{3/2}} \)
\( \frac{d}{dx} \left( \frac{5x^2}{\sqrt{x^2+3}} \right) = \frac{5x^3+30x}{(x^2+3)^{3/2}} \)
\( \frac{d}{dx} \left( \frac{5x^2}{\sqrt{x^2+3}} \right) = \frac{5x(x^2+6)}{(x^2+3)^{3/2}} \)
Therefore, the marginal cost function is: \( MC = \frac{5x(x^2+6)}{(x^2+3)^{3/2}} \).

(ii) Given cost function: \( C(x) = 7x + 130 \)
To find the marginal cost function (MC), we take the derivative of \( C(x) \) with respect to \( x \).
\( MC = \frac{dC}{dx} = \frac{d}{dx} (7x + 130) \)
\( MC = 7 \)
Since the marginal cost is a constant value (7), it is always constant, no matter how many units are produced. This means the cost of making one extra unit is always Rs. 7.
In simple words: For part (i), we found the marginal cost by taking the derivative of the total cost using the quotient rule. For part (ii), because the cost function is a simple straight line, the marginal cost (which is the slope of that line) is a constant number.

🎯 Exam Tip: For differentiation, know your rules: power rule, product rule, quotient rule, and chain rule. Always simplify your answer. When a cost function is linear (like `ax + b`), the marginal cost is simply `a` (the slope), which is constant.

Question 3.
(i) The cost function \( C(x) \) of a function is given by \( C(x) = 2x^2 – 4x + 5 \). Find the average cost, and the marginal cost when (a) \( x = 10 \) (b) when \( x = 2 \).
(ii) The cost function of a firm is given by \( C = 4x^2 – x + 70 \). Find the average cost and the marginal cost.
(iii) The cost function of a firm is given by \( C = 3x^2 – 2x + 3 \). Find (i) the average cost and (ii) the marginal cost when \( x = 3 \).
Answer:
(i) Given cost function: \( C(x) = 2x^2 – 4x + 5 \)
(a) Average Cost (AC) and Marginal Cost (MC) when \( x = 10 \):
First, find the general Average Cost function:
\( AC(x) = \frac{C(x)}{x} = \frac{2x^2 – 4x + 5}{x} = 2x – 4 + \frac{5}{x} \)
Now, calculate AC at \( x = 10 \):
\( AC(10) = 2(10) – 4 + \frac{5}{10} = 20 – 4 + 0.5 = 16 + 0.5 = 16.5 \)
Next, find the general Marginal Cost function:
\( MC(x) = \frac{dC}{dx} = \frac{d}{dx} (2x^2 – 4x + 5) = 4x – 4 \)
Now, calculate MC at \( x = 10 \):
\( MC(10) = 4(10) – 4 = 40 – 4 = 36 \)
(b) Average Cost (AC) and Marginal Cost (MC) when \( x = 2 \):
Using the general AC function from above:
\( AC(2) = 2(2) – 4 + \frac{5}{2} = 4 – 4 + 2.5 = 2.5 \)
Using the general MC function from above:
\( MC(2) = 4(2) – 4 = 8 – 4 = 4 \)

(ii) Given cost function: \( C(x) = 4x^2 – x + 70 \)
Average Cost (AC):
\( AC(x) = \frac{C(x)}{x} = \frac{4x^2 – x + 70}{x} = 4x – 1 + \frac{70}{x} \)
Marginal Cost (MC):
\( MC(x) = \frac{dC}{dx} = \frac{d}{dx} (4x^2 – x + 70) = 8x – 1 \)

(iii) Given cost function: \( C(x) = 3x^2 – 2x + 3 \)
Average Cost (AC) when \( x = 3 \):
First, find the general Average Cost function:
\( AC(x) = \frac{C(x)}{x} = \frac{3x^2 – 2x + 3}{x} = 3x – 2 + \frac{3}{x} \)
Now, calculate AC at \( x = 3 \):
\( AC(3) = 3(3) – 2 + \frac{3}{3} = 9 – 2 + 1 = 8 \)
Marginal Cost (MC) when \( x = 3 \):
First, find the general Marginal Cost function:
\( MC(x) = \frac{dC}{dx} = \frac{d}{dx} (3x^2 – 2x + 3) = 6x – 2 \)
Now, calculate MC at \( x = 3 \):
\( MC(3) = 6(3) – 2 = 18 – 2 = 16 \)
In simple words: For each cost function, we first find the formulas for average cost (total cost divided by items) and marginal cost (derivative of total cost). Then, we simply put the given number of items (x value) into these formulas to find the exact cost.

🎯 Exam Tip: Always present the general functions for AC and MC first, then substitute the specific values of `x` to avoid calculation errors and to show your complete understanding of the concepts.

Question 4. The total cost \( C(x) \), associated with production and making of x units of an item is given by \( C(x) = 0.005x^3 – 0.02x^2 + 30x + 5000 \); find
(i) the average cost function,
(ii) the average cost of output of 10 units,
(iii) the marginal cost function, and
(iv) the marginal cost when 3 units are produced.
Answer:
Given total cost function: \( C(x) = 0.005x^3 – 0.02x^2 + 30x + 5000 \)

(i) Average Cost Function (AC):
\( AC(x) = \frac{C(x)}{x} = \frac{0.005x^3 – 0.02x^2 + 30x + 5000}{x} \)
\( AC(x) = 0.005x^2 – 0.02x + 30 + \frac{5000}{x} \)

(ii) Average Cost of output of 10 units:
Substitute \( x = 10 \) into the AC function:
\( AC(10) = 0.005(10)^2 – 0.02(10) + 30 + \frac{5000}{10} \)
\( AC(10) = 0.005(100) – 0.2 + 30 + 500 \)
\( AC(10) = 0.5 – 0.2 + 30 + 500 \)
\( AC(10) = 530.3 \)

(iii) Marginal Cost Function (MC):
We find this by taking the derivative of \( C(x) \) with respect to \( x \).
\( MC(x) = \frac{dC}{dx} = \frac{d}{dx} (0.005x^3 – 0.02x^2 + 30x + 5000) \)
\( MC(x) = 0.005(3x^2) – 0.02(2x) + 30 \)
\( MC(x) = 0.015x^2 – 0.04x + 30 \)

(iv) Marginal Cost when 3 units are produced:
Substitute \( x = 3 \) into the MC function:
\( MC(3) = 0.015(3)^2 – 0.04(3) + 30 \)
\( MC(3) = 0.015(9) – 0.12 + 30 \)
\( MC(3) = 0.135 – 0.12 + 30 \)
\( MC(3) = 30.015 \)
In simple words: We calculated the average cost by dividing total cost by quantity. Then, we found the average cost for 10 units by putting 10 into that formula. For marginal cost, we took the derivative of the total cost. Finally, we found the marginal cost for 3 units by putting 3 into that formula.

🎯 Exam Tip: Pay close attention to decimal places during calculations to avoid small errors. Clearly label each part of your solution for readability.

Question 5.
(i) If a manufacturer's total cost function \( C \) is given by \( C = \frac{x^2}{25}+2 x \), find (a) the average cost function, (b) the marginal cost function, and (c) the marginal cost when 5 units are produced. Also, interpret the result.
(ii) The average cost function for a product is given by \( AC = 0.0002x^3 – 0.05x + 7 + \frac{8000}{x} \). Find the marginal cost function. What is the marginal cost when 100 units are produced. Interpret your result.
Answer:
(i) Given total cost function: \( C(x) = \frac{x^2}{25} + 2x \)
(a) Average Cost Function (AC):
\( AC(x) = \frac{C(x)}{x} = \frac{\frac{x^2}{25} + 2x}{x} \)
\( AC(x) = \frac{x}{25} + 2 \)
(b) Marginal Cost Function (MC):
\( MC(x) = \frac{dC}{dx} = \frac{d}{dx} \left( \frac{x^2}{25} + 2x \right) \)
\( MC(x) = \frac{2x}{25} + 2 \)
(c) Marginal Cost when 5 units are produced:
Substitute \( x = 5 \) into the MC function:
\( MC(5) = \frac{2(5)}{25} + 2 \)
\( MC(5) = \frac{10}{25} + 2 \)
\( MC(5) = \frac{2}{5} + 2 = 0.4 + 2 = 2.4 \)
Interpretation: When production increases by 1 unit (from 5 to 6 units), the additional cost incurred is approximately 2.4 units of currency. This means making the 6th unit will cost about Rs. 2.4.

(ii) Given average cost function: \( AC(x) = 0.0002x^3 – 0.05x + 7 + \frac{8000}{x} \)
First, find the total cost function \( C(x) \). We know that \( AC(x) = \frac{C(x)}{x} \), so \( C(x) = x \cdot AC(x) \).
\( C(x) = x \left( 0.0002x^3 – 0.05x + 7 + \frac{8000}{x} \right) \)
\( C(x) = 0.0002x^4 – 0.05x^2 + 7x + 8000 \)
Now, find the Marginal Cost Function (MC) by differentiating \( C(x) \):
\( MC(x) = \frac{dC}{dx} = \frac{d}{dx} (0.0002x^4 – 0.05x^2 + 7x + 8000) \)
\( MC(x) = 0.0002(4x^3) – 0.05(2x) + 7 \)
\( MC(x) = 0.0008x^3 – 0.1x + 7 \)
Marginal Cost when 100 units are produced:
Substitute \( x = 100 \) into the MC function:
\( MC(100) = 0.0008(100)^3 – 0.1(100) + 7 \)
\( MC(100) = 0.0008(1,000,000) – 10 + 7 \)
\( MC(100) = 800 – 10 + 7 \)
\( MC(100) = 797 \)
Interpretation: When production increases by 1 unit (from 100 to 101 units), the additional cost incurred is approximately 797 units of currency. This means making the 101st unit will cost about Rs. 797.
In simple words: For the first part, we found the average and marginal cost formulas. Then we calculated the marginal cost for 5 units and explained that it's the extra cost for one more item. For the second part, we first found the total cost from the average cost, then found the marginal cost formula, and finally calculated the marginal cost for 100 units, explaining what that number means.

🎯 Exam Tip: When given AC and asked for MC, always remember to first derive the total cost function \( C(x) \) by multiplying \( AC(x) \) by \( x \), then differentiate \( C(x) \) to get \( MC(x) \). Interpreting the marginal cost means explaining what the calculated value represents in terms of additional cost per unit.

Question 6.
(i) The total cost function for a production and marketing activity is given by \( C(x) = \frac { 3 }{ 4 }x^2 – 7x + 27 \). Find the level of output (number of units produced) for which MC = AC.
(ii) If \( C(x) = 0.05x – 0.2x^2 – 5 \), find the level of output x for which the average cost function AC becomes equal to the marginal cost.
Answer:
(i) Given total cost function: \( C(x) = \frac{3}{4}x^2 – 7x + 27 \)
First, find the Marginal Cost (MC) function:
\( MC(x) = \frac{dC}{dx} = \frac{d}{dx} \left( \frac{3}{4}x^2 – 7x + 27 \right) \)
\( MC(x) = \frac{3}{4}(2x) – 7 \)
\( MC(x) = \frac{3}{2}x – 7 \)
Next, find the Average Cost (AC) function:
\( AC(x) = \frac{C(x)}{x} = \frac{\frac{3}{4}x^2 – 7x + 27}{x} \)
\( AC(x) = \frac{3}{4}x – 7 + \frac{27}{x} \)
Now, set \( MC(x) = AC(x) \) and solve for \( x \):
\( \frac{3}{2}x – 7 = \frac{3}{4}x – 7 + \frac{27}{x} \)
Subtract 7 from both sides:
\( \frac{3}{2}x = \frac{3}{4}x + \frac{27}{x} \)
Bring all \( x \) terms to one side:
\( \frac{3}{2}x – \frac{3}{4}x = \frac{27}{x} \)
Find a common denominator for the \( x \) terms (4):
\( \frac{6x – 3x}{4} = \frac{27}{x} \)
\( \frac{3x}{4} = \frac{27}{x} \)
Cross-multiply:
\( 3x \cdot x = 27 \cdot 4 \)
\( 3x^2 = 108 \)
\( x^2 = \frac{108}{3} \)
\( x^2 = 36 \)
\( x = \pm 6 \)
Since the number of units produced (\( x \)) cannot be negative, we take \( x = 6 \).
So, the level of output for which MC = AC is 6 units.

(ii) Given total cost function: \( C(x) = 0.05x – 0.2x^2 – 5 \)
First, find the Marginal Cost (MC) function:
\( MC(x) = \frac{dC}{dx} = \frac{d}{dx} (0.05x – 0.2x^2 – 5) \)
\( MC(x) = 0.05 – 0.4x \)
Next, find the Average Cost (AC) function:
\( AC(x) = \frac{C(x)}{x} = \frac{0.05x – 0.2x^2 – 5}{x} \)
\( AC(x) = 0.05 – 0.2x – \frac{5}{x} \)
Now, set \( AC(x) = MC(x) \) and solve for \( x \):
\( 0.05 – 0.2x – \frac{5}{x} = 0.05 – 0.4x \)
Subtract 0.05 from both sides:
\( -0.2x – \frac{5}{x} = -0.4x \)
Add \( 0.4x \) to both sides:
\( 0.4x – 0.2x = \frac{5}{x} \)
\( 0.2x = \frac{5}{x} \)
Multiply both sides by \( x \):
\( 0.2x^2 = 5 \)
\( x^2 = \frac{5}{0.2} \)
\( x^2 = 25 \)
\( x = \pm 5 \)
Since the number of units produced (\( x \)) must be positive, we take \( x = 5 \).
So, the level of output for which AC = MC is 5 units.
In simple words: For both parts, we first found the formulas for marginal cost (MC) and average cost (AC). Then, we set these two formulas equal to each other and solved for 'x' to find the number of units where their costs are the same.

🎯 Exam Tip: When solving for `x` where MC = AC, remember that output `x` cannot be negative. Always discard negative solutions in such contexts. This point is often overlooked and can lead to incorrect answers.

Question 7. The total cost function is given by \( C = x + 2x^3 – 3.5x^2 \), find the marginal average cost function (MAC). Also, find the points where the MC curve cuts the x-axis and y-axis.
Answer:
Given total cost function: \( C(x) = x + 2x^3 – 3.5x^2 \)
First, find the Average Cost (AC) function:
\( AC(x) = \frac{C(x)}{x} = \frac{x + 2x^3 – 3.5x^2}{x} \)
\( AC(x) = 1 + 2x^2 – 3.5x \)
Now, find the Marginal Average Cost (MAC) function. This is the derivative of the AC function.
\( MAC(x) = \frac{d}{dx}(AC(x)) = \frac{d}{dx} (1 + 2x^2 – 3.5x) \)
\( MAC(x) = 4x – 3.5 \)

Next, we need to find the Marginal Cost (MC) function. This is the derivative of the total cost function \( C(x) \).
\( MC(x) = \frac{dC}{dx} = \frac{d}{dx} (x + 2x^3 – 3.5x^2) \)
\( MC(x) = 1 + 2(3x^2) – 3.5(2x) \)
\( MC(x) = 1 + 6x^2 – 7x \)

Now, find where the MC curve cuts the x-axis and y-axis:
To find where MC cuts the x-axis, set \( MC(x) = 0 \):
\( 1 + 6x^2 – 7x = 0 \)
Rearrange into standard quadratic form: \( 6x^2 – 7x + 1 = 0 \)
Factor the quadratic equation: We look for two numbers that multiply to 6 (6*1) and add to -7 (-6 and -1).
\( 6x^2 – 6x – x + 1 = 0 \)
\( 6x(x – 1) – 1(x – 1) = 0 \)
\( (x – 1)(6x – 1) = 0 \)
This gives two possible values for \( x \):
\( x – 1 = 0 \implies x = 1 \)
\( 6x – 1 = 0 \implies x = \frac{1}{6} \)
So, the MC curve cuts the x-axis at points \( (1, 0) \) and \( \left( \frac{1}{6}, 0 \right) \).

To find where MC cuts the y-axis, set \( x = 0 \):
\( MC(0) = 1 + 6(0)^2 – 7(0) \)
\( MC(0) = 1 \)
So, the MC curve cuts the y-axis at point \( (0, 1) \).
In simple words: First, we found the average cost by dividing total cost by 'x', and then the marginal average cost by taking the derivative of the average cost. Next, we found the marginal cost by taking the derivative of the total cost. Finally, to see where the marginal cost line crosses the x-axis, we set it to zero and solved for 'x'. To see where it crosses the y-axis, we set 'x' to zero.

🎯 Exam Tip: Remember to distinguish between Marginal Cost (MC) and Marginal Average Cost (MAC). MC is the derivative of Total Cost, while MAC is the derivative of Average Cost. For intercepts, set the function to zero for x-intercepts, and set x to zero for the y-intercept.

Question 8. Verify that \( \frac { d }{ dx } (AC) = \frac { 1 }{ x }(MC – AC) \), given that
(i) \( C = 3 – 2x + 5x^2 \)
(ii) \( C= a + bx + cx^2 \)
(iii) \( C(x) = ax^3 + bx^2 + cx + d \)
Answer:
(i) Given total cost function: \( C = 3 – 2x + 5x^2 \)
First, find Average Cost (AC) and Marginal Cost (MC):
\( AC = \frac{C}{x} = \frac{3 – 2x + 5x^2}{x} = \frac{3}{x} – 2 + 5x \)
\( MC = \frac{dC}{dx} = \frac{d}{dx} (3 – 2x + 5x^2) = -2 + 10x \)

Now, calculate the Left Hand Side (LHS):
\( LHS = \frac{d}{dx}(AC) = \frac{d}{dx} \left( \frac{3}{x} – 2 + 5x \right) = -3x^{-2} + 5 = -\frac{3}{x^2} + 5 \)

Next, calculate the Right Hand Side (RHS):
\( RHS = \frac{1}{x}(MC – AC) = \frac{1}{x} \left( (-2 + 10x) – \left(\frac{3}{x} – 2 + 5x\right) \right) \)
\( RHS = \frac{1}{x} \left( -2 + 10x – \frac{3}{x} + 2 – 5x \right) \)
\( RHS = \frac{1}{x} \left( 5x – \frac{3}{x} \right) \)
\( RHS = 5 – \frac{3}{x^2} \)
Since LHS = RHS, the identity is verified for \( C = 3 – 2x + 5x^2 \).

(ii) Given total cost function: \( C = a + bx + cx^2 \)
First, find Average Cost (AC) and Marginal Cost (MC):
\( AC = \frac{C}{x} = \frac{a + bx + cx^2}{x} = \frac{a}{x} + b + cx \)
\( MC = \frac{dC}{dx} = \frac{d}{dx} (a + bx + cx^2) = b + 2cx \)

Now, calculate the Left Hand Side (LHS):
\( LHS = \frac{d}{dx}(AC) = \frac{d}{dx} \left( \frac{a}{x} + b + cx \right) = -ax^{-2} + c = -\frac{a}{x^2} + c \)

Next, calculate the Right Hand Side (RHS):
\( RHS = \frac{1}{x}(MC – AC) = \frac{1}{x} \left( (b + 2cx) – \left(\frac{a}{x} + b + cx\right) \right) \)
\( RHS = \frac{1}{x} \left( b + 2cx – \frac{a}{x} – b – cx \right) \)
\( RHS = \frac{1}{x} \left( cx – \frac{a}{x} \right) \)
\( RHS = c – \frac{a}{x^2} \)
Since LHS = RHS, the identity is verified for \( C = a + bx + cx^2 \).

(iii) Given total cost function: \( C(x) = ax^3 + bx^2 + cx + d \)
First, find Average Cost (AC) and Marginal Cost (MC):
\( AC(x) = \frac{C(x)}{x} = \frac{ax^3 + bx^2 + cx + d}{x} = ax^2 + bx + c + \frac{d}{x} \)
\( MC(x) = \frac{dC}{dx} = \frac{d}{dx} (ax^3 + bx^2 + cx + d) = 3ax^2 + 2bx + c \)

Now, calculate the Left Hand Side (LHS):
\( LHS = \frac{d}{dx}(AC) = \frac{d}{dx} \left( ax^2 + bx + c + \frac{d}{x} \right) = 2ax + b – dx^{-2} = 2ax + b – \frac{d}{x^2} \)

Next, calculate the Right Hand Side (RHS):
\( RHS = \frac{1}{x}(MC – AC) = \frac{1}{x} \left( (3ax^2 + 2bx + c) – \left(ax^2 + bx + c + \frac{d}{x}\right) \right) \)
\( RHS = \frac{1}{x} \left( 3ax^2 + 2bx + c – ax^2 – bx – c – \frac{d}{x} \right) \)
\( RHS = \frac{1}{x} \left( 2ax^2 + bx – \frac{d}{x} \right) \)
\( RHS = 2ax + b – \frac{d}{x^2} \)
Since LHS = RHS, the identity is verified for \( C(x) = ax^3 + bx^2 + cx + d \).
In simple words: For each cost function, we first found the average cost (AC) and marginal cost (MC). Then, we calculated both sides of the given equation separately. Since both sides ended up being the same, the identity was proven correct for all three cost functions.

🎯 Exam Tip: When verifying identities, clearly separate your LHS and RHS calculations. Take extra care with algebraic manipulation, especially when dealing with negative signs and fractions, to avoid errors. This identity is important as it links average and marginal costs.

Question 9. The average cost of producing x units of a commodity is given by the equation \( AC = \frac{x^2}{200} – \frac{x}{50} – 30 + \frac{5000}{x} \). Find the marginal cost (MC) function and verify that \( \frac{d AC}{d x} = \frac{M C-A C}{x} \).
Answer:
Given average cost function: \( AC(x) = \frac{x^2}{200} – \frac{x}{50} – 30 + \frac{5000}{x} \)
First, find the total cost function \( C(x) \). We know \( C(x) = x \cdot AC(x) \).
\( C(x) = x \left( \frac{x^2}{200} – \frac{x}{50} – 30 + \frac{5000}{x} \right) \)
\( C(x) = \frac{x^3}{200} – \frac{x^2}{50} – 30x + 5000 \)

Next, find the Marginal Cost (MC) function by differentiating \( C(x) \) with respect to \( x \):
\( MC(x) = \frac{dC}{dx} = \frac{d}{dx} \left( \frac{x^3}{200} – \frac{x^2}{50} – 30x + 5000 \right) \)
\( MC(x) = \frac{3x^2}{200} – \frac{2x}{50} – 30 \)

Now, we need to verify the given identity \( \frac{d AC}{d x} = \frac{M C-A C}{x} \).
Calculate the Left Hand Side (LHS):
\( LHS = \frac{d}{dx}(AC) = \frac{d}{dx} \left( \frac{x^2}{200} – \frac{x}{50} – 30 + \frac{5000}{x} \right) \)
\( LHS = \frac{2x}{200} – \frac{1}{50} – 5000x^{-2} \)
\( LHS = \frac{x}{100} – \frac{1}{50} – \frac{5000}{x^2} \)

Calculate the Right Hand Side (RHS):
\( RHS = \frac{1}{x}(MC – AC) \)
\( RHS = \frac{1}{x} \left[ \left( \frac{3x^2}{200} – \frac{2x}{50} – 30 \right) – \left( \frac{x^2}{200} – \frac{x}{50} – 30 + \frac{5000}{x} \right) \right] \)
\( RHS = \frac{1}{x} \left[ \frac{3x^2}{200} – \frac{2x}{50} – 30 – \frac{x^2}{200} + \frac{x}{50} + 30 – \frac{5000}{x} \right] \)
\( RHS = \frac{1}{x} \left[ \left( \frac{3x^2}{200} – \frac{x^2}{200} \right) + \left( -\frac{2x}{50} + \frac{x}{50} \right) – \frac{5000}{x} \right] \)
\( RHS = \frac{1}{x} \left[ \frac{2x^2}{200} – \frac{x}{50} – \frac{5000}{x} \right] \)
\( RHS = \frac{1}{x} \left[ \frac{x^2}{100} – \frac{x}{50} – \frac{5000}{x} \right] \)
\( RHS = \frac{x}{100} – \frac{1}{50} – \frac{5000}{x^2} \)
Since LHS = RHS, the identity is verified. This identity shows the relationship between marginal cost, average cost, and the rate of change of average cost.
In simple words: First, we used the average cost to find the total cost, and then we found the marginal cost by taking the derivative. After that, we calculated both sides of the given equation to make sure they were equal, which confirmed the relationship between average and marginal costs.

🎯 Exam Tip: The identity \( \frac{d AC}{d x} = \frac{M C-A C}{x} \) is a fundamental relationship in economics. Ensure all your algebraic steps are clear and correct during verification. It indicates that if MC > AC, average cost is rising, and if MC < AC, average cost is falling.

Question 10. The total cost of production of x units of a certain product is given by \( C(x) = 2x\left(\frac{x+4}{x+1}\right)+6 \). Show that the marginal cost falls continuously as the output x increases.
Answer:
Given total cost function: \( C(x) = 2x\left(\frac{x+4}{x+1}\right)+6 \)
First, simplify \( C(x) \):
\( C(x) = \frac{2x(x+4)}{x+1}+6 = \frac{2x^2+8x}{x+1}+6 \)
To show that the marginal cost falls continuously as output \( x \) increases, we need to find the marginal cost function (MC) and then show that its derivative \( \frac{d}{dx}(MC) \) is always negative.

Find the Marginal Cost (MC) function by differentiating \( C(x) \) with respect to \( x \):
\( MC = \frac{dC}{dx} = \frac{d}{dx} \left( \frac{2x^2+8x}{x+1} + 6 \right) \)
The derivative of 6 is 0.
For the first term, use the quotient rule: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \).
Let \( u = 2x^2+8x \) and \( v = x+1 \).
Then \( u' = 4x+8 \) and \( v' = 1 \).
\( MC = \frac{(4x+8)(x+1) - (2x^2+8x)(1)}{(x+1)^2} \)
\( MC = \frac{4x^2+4x+8x+8 - 2x^2-8x}{(x+1)^2} \)
\( MC = \frac{2x^2+4x+8}{(x+1)^2} \)
This is the Marginal Cost function. We can factor out a 2 from the numerator:
\( MC = \frac{2(x^2+2x+4)}{(x+1)^2} \)
We can rewrite \( x^2+2x+4 \) as \( (x+1)^2+3 \).
\( MC = \frac{2((x+1)^2+3)}{(x+1)^2} = 2 \left( 1 + \frac{3}{(x+1)^2} \right) = 2 + \frac{6}{(x+1)^2} \)

Now, find the derivative of the Marginal Cost (MC) function to check its trend:
\( \frac{d}{dx}(MC) = \frac{d}{dx} \left( 2 + 6(x+1)^{-2} \right) \)
\( \frac{d}{dx}(MC) = 0 + 6(-2)(x+1)^{-3}(1) \)
\( \frac{d}{dx}(MC) = -12(x+1)^{-3} = \frac{-12}{(x+1)^3} \)
For \( x > 0 \) (since \( x \) represents units of output), \( (x+1) \) will always be positive, so \( (x+1)^3 \) will also be positive. Therefore, \( \frac{-12}{(x+1)^3} \) will always be negative.
Since \( \frac{d}{dx}(MC) < 0 \) for all \( x > 0 \), the marginal cost falls continuously as the output \( x \) increases.
In simple words: We first found the marginal cost by taking the derivative of the total cost. Then, to prove that marginal cost always decreases as more items are made, we took the derivative of the marginal cost itself. Since this second derivative was always a negative number, it confirms that marginal cost continuously falls.

🎯 Exam Tip: To prove that a function is continuously increasing or decreasing, you must show that its first derivative is always positive or always negative, respectively, over the relevant domain (usually \( x > 0 \) for economic contexts). Clearly show the differentiation steps for both the cost function and its derivative.

Question 11. Verify for the cost functions \( C(x) = ax\left(\frac{x+b}{x+c}\right) + d \), where \( a, b, c, d > 0 \) and \( b > c \), that the average and marginal cost curves fall continuously with increasing output.
Answer:
Given cost function: \( C(x) = ax\left(\frac{x+b}{x+c}\right) + d \)
We are given that \( a, b, c, d > 0 \) and \( b > c \).

First, let's analyze the Average Cost (AC) curve:
\( AC(x) = \frac{C(x)}{x} = \frac{ax\left(\frac{x+b}{x+c}\right) + d}{x} \)
\( AC(x) = a\left(\frac{x+b}{x+c}\right) + \frac{d}{x} \)
To determine if AC falls continuously, we need to find \( \frac{d}{dx}(AC) \) and check if it's negative.
\( \frac{d}{dx}(AC) = \frac{d}{dx} \left( a\left(\frac{x+b}{x+c}\right) + \frac{d}{x} \right) \)
\( \frac{d}{dx}(AC) = a \cdot \frac{d}{dx}\left(\frac{x+b}{x+c}\right) + \frac{d}{dx}\left(\frac{d}{x}\right) \)
For \( \frac{d}{dx}\left(\frac{x+b}{x+c}\right) \), use the quotient rule: \( \frac{1(x+c) - (x+b)1}{(x+c)^2} = \frac{x+c-x-b}{(x+c)^2} = \frac{c-b}{(x+c)^2} \)
And \( \frac{d}{dx}\left(\frac{d}{x}\right) = -dx^{-2} = -\frac{d}{x^2} \).
So, \( \frac{d}{dx}(AC) = a \left( \frac{c-b}{(x+c)^2} \right) – \frac{d}{x^2} \)
We are given \( a > 0, d > 0, x > 0 \). Also, \( b > c \implies c – b < 0 \).
Therefore, \( a \left( \frac{c-b}{(x+c)^2} \right) \) is a negative term.
And \( -\frac{d}{x^2} \) is also a negative term.
Since both parts are negative, their sum \( \frac{d}{dx}(AC) \) is always negative.
Thus, the average cost curve falls continuously with increasing output.

Next, let's analyze the Marginal Cost (MC) curve:
First, simplify \( C(x) \):
\( C(x) = a \frac{x^2+bx}{x+c} + d \)
Find the Marginal Cost (MC) function by differentiating \( C(x) \) with respect to \( x \):
\( MC = \frac{dC}{dx} = \frac{d}{dx} \left( a \frac{x^2+bx}{x+c} + d \right) \)
\( MC = a \cdot \frac{d}{dx}\left(\frac{x^2+bx}{x+c}\right) + 0 \)
For \( \frac{d}{dx}\left(\frac{x^2+bx}{x+c}\right) \), use the quotient rule.
Let \( u = x^2+bx \) and \( v = x+c \).
Then \( u' = 2x+b \) and \( v' = 1 \).
\( \frac{d}{dx}\left(\frac{x^2+bx}{x+c}\right) = \frac{(2x+b)(x+c) - (x^2+bx)(1)}{(x+c)^2} \)
\( = \frac{2x^2+2cx+bx+bc - x^2-bx}{(x+c)^2} \)
\( = \frac{x^2+2cx+bc}{(x+c)^2} \)
So, \( MC(x) = a \frac{x^2+2cx+bc}{(x+c)^2} \)

Now, find \( \frac{d}{dx}(MC) \) to determine if MC falls continuously:
\( \frac{d}{dx}(MC) = a \cdot \frac{d}{dx} \left( \frac{x^2+2cx+bc}{(x+c)^2} \right) \)
Again, use the quotient rule for \( \frac{d}{dx} \left( \frac{u}{v} \right) \).
Let \( u = x^2+2cx+bc \) and \( v = (x+c)^2 \).
Then \( u' = 2x+2c \).
And \( v' = 2(x+c)(1) = 2(x+c) \).
\( \frac{d}{dx} \left( \frac{x^2+2cx+bc}{(x+c)^2} \right) = \frac{(2x+2c)(x+c)^2 - (x^2+2cx+bc)2(x+c)}{((x+c)^2)^2} \)
Factor out \( (x+c) \) from the numerator:
\( = \frac{(x+c) [ (2x+2c)(x+c) - 2(x^2+2cx+bc) ]}{(x+c)^4} \)
\( = \frac{(2x+2c)(x+c) - 2(x^2+2cx+bc)}{(x+c)^3} \)
\( = \frac{2(x+c)(x+c) - 2(x^2+2cx+bc)}{(x+c)^3} \)
\( = \frac{2(x^2+2cx+c^2) - 2(x^2+2cx+bc)}{(x+c)^3} \)
\( = \frac{2x^2+4cx+2c^2 - 2x^2-4cx-2bc}{(x+c)^3} \)
\( = \frac{2c^2 - 2bc}{(x+c)^3} = \frac{2c(c-b)}{(x+c)^3} \)
So, \( \frac{d}{dx}(MC) = a \cdot \frac{2c(c-b)}{(x+c)^3} \)
We are given \( a > 0, c > 0 \). Also, \( b > c \implies c – b < 0 \). And for \( x > 0 \), \( (x+c)^3 > 0 \).
Therefore, \( a \cdot \frac{2c(c-b)}{(x+c)^3} \) is a product of positive numbers \( a, 2c, (x+c)^3 \) and a negative number \( (c-b) \).
So, \( \frac{d}{dx}(MC) \) is always negative.
Thus, both the average cost and marginal cost curves fall continuously with increasing output under the given conditions.
In simple words: We first found the formulas for average cost (AC) and marginal cost (MC). Then, we took the derivative of both AC and MC. Because the derivatives of both AC and MC turned out to be negative, it means both costs continuously decrease as more items are produced. This happens because 'b' is greater than 'c' in the cost function.

🎯 Exam Tip: When proving a function continuously falls, ensure every term in the derivative is accounted for and its sign is correctly determined based on the given conditions (e.g., \( a,b,c,d > 0 \) and \( b > c \)). Clearly state the conclusion for both AC and MC after their derivatives are analyzed.

Question 12. The average cost function associated with producing and marketing x units of an item is given by \( AC = 2x – 11 + \frac{50}{x} \). Find the total cost function and marginal cost function. Also, find the range of the output for which AC is increasing.
Answer:
Given average cost function: \( AC(x) = 2x – 11 + \frac{50}{x} \)

1. Total Cost Function \( C(x) \):
We know that \( AC(x) = \frac{C(x)}{x} \), so \( C(x) = x \cdot AC(x) \).
\( C(x) = x \left( 2x – 11 + \frac{50}{x} \right) \)
\( C(x) = 2x^2 – 11x + 50 \)

2. Marginal Cost Function \( MC(x) \):
\( MC(x) = \frac{dC}{dx} = \frac{d}{dx} (2x^2 – 11x + 50) \)
\( MC(x) = 4x – 11 \)

3. Range of output for which AC is increasing:
To find when AC is increasing, we need to find the derivative of AC, \( \frac{d}{dx}(AC) \), and set it greater than 0.
\( \frac{d}{dx}(AC) = \frac{d}{dx} \left( 2x – 11 + 50x^{-1} \right) \)
\( \frac{d}{dx}(AC) = 2 – 50x^{-2} \)
\( \frac{d}{dx}(AC) = 2 – \frac{50}{x^2} \)
For AC to be increasing, \( \frac{d}{dx}(AC) > 0 \):
\( 2 – \frac{50}{x^2} > 0 \)
\( 2 > \frac{50}{x^2} \)
Multiply both sides by \( x^2 \) (since \( x > 0 \), \( x^2 \) is positive, so the inequality direction does not change):
\( 2x^2 > 50 \)
\( x^2 > \frac{50}{2} \)
\( x^2 > 25 \)
Take the square root of both sides:
\( \sqrt{x^2} > \sqrt{25} \)
\( |x| > 5 \)
Since output \( x \) must be positive, the range for which AC is increasing is \( x > 5 \). This means when you produce more than 5 units, the average cost starts to go up.
In simple words: First, we used the average cost formula to find the total cost, and then we found the marginal cost by taking the derivative of the total cost. To find when the average cost is going up, we took the derivative of the average cost and set it to be greater than zero, then solved for 'x'.

🎯 Exam Tip: Remember that "increasing" implies a positive first derivative, and "decreasing" implies a negative first derivative. Always consider the practical domain of `x` (i.e., `x > 0` for output) when interpreting results in economic problems.

Question 13. The total cost function for x units is given by \( C(x) = \sqrt{6x+5}+2500 \). Show that the marginal cost decreases as the output x increases.
Answer:
Given total cost function: \( C(x) = \sqrt{6x+5}+2500 \)
To show that the marginal cost decreases as output \( x \) increases, we need to find the marginal cost function (MC) and then show that its derivative \( \frac{d}{dx}(MC) \) is always negative.

Find the Marginal Cost (MC) function by differentiating \( C(x) \) with respect to \( x \):
\( C(x) = (6x+5)^{1/2} + 2500 \)
\( MC = \frac{dC}{dx} = \frac{d}{dx} \left( (6x+5)^{1/2} + 2500 \right) \)
Using the chain rule for \( (6x+5)^{1/2} \): \( \frac{1}{2}(6x+5)^{-1/2} \cdot 6 = 3(6x+5)^{-1/2} \)
The derivative of 2500 is 0.
So, \( MC(x) = 3(6x+5)^{-1/2} = \frac{3}{\sqrt{6x+5}} \)

Now, find the derivative of the Marginal Cost (MC) function to check its trend:
\( \frac{d}{dx}(MC) = \frac{d}{dx} \left( 3(6x+5)^{-1/2} \right) \)
\( \frac{d}{dx}(MC) = 3 \cdot \left( -\frac{1}{2} \right) (6x+5)^{-3/2} \cdot 6 \)
\( \frac{d}{dx}(MC) = -9(6x+5)^{-3/2} \)
\( \frac{d}{dx}(MC) = \frac{-9}{(6x+5)^{3/2}} \)
For output \( x > 0 \), we know that \( 6x+5 \) will always be positive. Therefore, \( (6x+5)^{3/2} \) will also be positive.
Since the numerator is -9 (a negative number) and the denominator is always positive, the entire expression \( \frac{-9}{(6x+5)^{3/2}} \) will always be negative.
Thus, \( \frac{d}{dx}(MC) < 0 \) for all \( x > 0 \). This proves that the marginal cost decreases continuously as the output \( x \) increases.
In simple words: We first found the marginal cost by taking the derivative of the total cost function. Then, we took the derivative of the marginal cost itself. Since this second derivative was always a negative number, it shows that the marginal cost keeps going down as more products are made.

🎯 Exam Tip: To show a function is decreasing, its first derivative must be negative. Be careful with fractional exponents and the chain rule when differentiating square root functions. Always specify the domain (e.g., \( x > 0 \)) when drawing conclusions about increasing/decreasing behavior in economic contexts.