Maharashtra Board Class 12 Maths Part 1 Chapter 4 Applications of Derivatives 4.4 Solutions

Question 1. The demand function of a commodity at price P is given as D = 40 - \( \frac{5P}{8} \). Check whether it is an increasing or decreasing function.
Answer: Solution: D = 40 - \( \frac{5P}{8} \)
\( \therefore \frac{dD}{dP} = \frac{d}{dP} (40 - \frac{5P}{8}) \)
\( = 0 - \frac{5}{8} \times 1 \)
\( = - \frac{5}{8} \) Hence, the given function is decreasing function.
In simple words: The demand function D = 40 - 5P/8 is decreasing because its derivative with respect to price (dP/dD) is -5/8, which is a negative value. A negative derivative indicates that as the price increases, the demand decreases.

🎯 Exam Tip: When analyzing if a function is increasing or decreasing, always calculate its first derivative. A positive derivative indicates an increasing function, while a negative derivative indicates a decreasing function.

Question 2. The price P for demand D is given as P = 183 + 120D - 3D², find D for which price is increasing.
Answer: Solution: P = 183 + 120D - 3D²
\( \therefore \frac{dP}{dD} = \frac{d}{dD} (183 + 120D - 3D^2) \)
\( = 0 + 120 \times 1 - 3 \times 2D \)
\( = 120 - 6D \) If price P is increasing, then \( \frac{dP}{dD} > 0 \)
\( \therefore 120 - 6D > 0 \)
\( \therefore 120 > 6D \)
\( \therefore D < 20 \) Hence, the price is increasing when D < 20.
In simple words: To find when the price P is increasing, we take the derivative of P with respect to D, which is 120 - 6D. For P to be increasing, this derivative must be positive. Solving 120 - 6D > 0 gives D < 20.

🎯 Exam Tip: Remember that for a function to be increasing, its first derivative must be greater than zero. For a function to be decreasing, its first derivative must be less than zero. Always show the inequality step clearly.

Question 3. The total cost function for production of x articles is given as C = 100 + 600x - 3x². Find the values of x for which the total cost is decreasing.
Answer: Solution: The cost function is given as C = 100 + 600x - 3x²
\( \therefore \frac{dC}{dx} = \frac{d}{dx} (100 + 600x - 3x^2) \)
\( = 0 + 600 \times 1 - 3 \times 2x \)
\( = 600 - 6x \) If the total cost is decreasing, then \( \frac{dC}{dx} < 0 \)
\( \therefore 600 - 6x < 0 \)
\( \therefore 600 < 6x \)
\( \therefore x > 100 \) Hence, the total cost is decreasing for x > 100.
In simple words: The total cost decreases when the marginal cost (the derivative of the total cost function) is negative. We found the marginal cost to be 600 - 6x, and setting it less than zero shows that the total cost decreases when x is greater than 100.

🎯 Exam Tip: Marginal cost is the derivative of the total cost function. For total cost to be decreasing, marginal cost should be negative. Ensure correct differentiation and inequality solving for full marks.

Question 4. The manufacturing company produces x items at the total cost of Rs.(180 + 4x). The demand function for this product is P = (240 - x). Find x for which
(i) revenue is increasing
(ii) profit is increasing.
Answer: Solution:
(i) Let R be the total revenue. Then R = P.x = (240 - x)x
\( \therefore R = 240x - x^2 \)
\( \therefore \frac{dR}{dx} = \frac{d}{dx} (240x - x^2) \)
\( = 240 \times 1 - 2x \)
\( = 240 - 2x \) R is increasing, if \( \frac{dR}{dx} > 0 \) i.e. if 240 - 2x > 0 i.e. if 240 > 2x i.e. if x < 120 Hence, the revenue is increasing, if x < 120.
(ii) Profit \( \pi = R-C \)
\( \therefore \pi = (240x - x^2) - (180 + 4x) \)
\( = 240x - x^2 - 180 - 4x \)
\( = 236x - x^2 - 180 \)
\( \therefore \frac{d\pi}{dx} = \frac{d}{dx} (236x - x^2 - 180) \)
\( = 236 \times 1 - 2x - 0 \)
\( = 236 - 2x \) Profit is increasing, if \( \frac{d\pi}{dx} > 0 \) i.e. if 236 - 2x > 0 i.e. if 236 > 2x i.e. if x < 118 Hence, the profit is increasing, if x < 118.
In simple words: Revenue increases when marginal revenue (dR/dx) is positive, which occurs for x < 120. Profit increases when marginal profit (dπ/dx) is positive, occurring for x < 118. Both calculations involve finding the derivative and setting it greater than zero.

🎯 Exam Tip: For revenue and profit problems, clearly define the revenue (R=P.x) and profit (π=R-C) functions first. Then, apply differentiation to find marginal revenue and marginal profit, using inequalities to determine increasing/decreasing ranges.

Question 5. For manufacturing x units, labour cost is 150 - 54x and processing cost is x². Price of each unit is p = 10800 - 4x². Find the values of x for which
(i) total cost is decreasing
(ii) revenue is increasing.
Answer: Solution:
(i) Total cost C = labour cost + processing cost
\( \therefore C = 150 - 54x + x^2 \)
\( \therefore \frac{dC}{dx} = \frac{d}{dx} (150 - 54x + x^2) \)
\( = 0 - 54 \times 1 + 2x \)
\( = -54 + 2x \) The total cost is decreasing, if \( \frac{dC}{dx} < 0 \) i.e. if -54 + 2x < 0 i.e. if 2x < 54 i.e. if x < 27 Hence, the total cost is decreasing, if x < 27.
(ii) The total revenue R is given as R = p.x \( R = (10800 - 4x^2) x \) \( R = 10800x - 4x^3 \)
\( \therefore \frac{dR}{dx} = \frac{d}{dx} (10800x - 4x^3) \)
\( = 10800 \times 1 - 4 \times 3x^2 \)
\( = 10800 - 12x^2 \) The revenue is increasing, if \( \frac{dR}{dx} > 0 \) i.e. if 10800 - 12x² > 0 i.e. if 10800 > 12x² i.e. if x² < 900 i.e. if x < 30 ......[ \( \because \) x > 0] Hence, the revenue is increasing, if x < 30.
In simple words: For the total cost to decrease, its derivative (marginal cost) must be negative, leading to x < 27. For revenue to increase, its derivative (marginal revenue) must be positive, which means x < 30.

🎯 Exam Tip: When dealing with cost and revenue functions, remember to define total cost (C) and total revenue (R) correctly based on the given information before computing derivatives. Pay attention to the domain of x (x > 0 for production quantity).

Question 6. The total cost of manufacturing x articles is C = 47x + 300x² - x⁴. Find x, for which average cost is
(i) increasing
(ii) decreasing.
Answer: Solution: The total cost is given as C = 47x + 300x² - x⁴
\( \therefore \) the average cost is given by \( C_A = \frac{C}{x} = \frac{47x+300x^2-x^4}{x} \)
\( \therefore C_A = 47+300x - x^3 \)
\( \therefore \frac{dC_A}{dx} = \frac{d}{dx} (47+300x-x^3) \)
\( = 0+300 \times 1 - 3x^2 = 300-3x^2 \)
(i) CA is increasing, if \( \frac{dC_A}{dx} > 0 \) i.e. if 300 - 3x² > 0 i.e. if 300 > 3x² i.e. if x² < 100 i.e. if x < 10 .....[ \( \because \) x > 0] Hence, the average cost is increasing, if x < 10.
(ii) CA is decreasing, if \( \frac{dC_A}{dx} < 0 \) i.e. if 300 - 3x² < 0 i.e. if 300 < 3x² i.e. if x² > 100 i.e. if x > 10 ......[ \( \because \) x > 0] Hence, the average cost is decreasing, if x > 10.
In simple words: First, calculate the average cost function (C_A = C/x). Then, find its derivative (dC_A/dx). The average cost is increasing when this derivative is positive (x < 10) and decreasing when it's negative (x > 10).

🎯 Exam Tip: Average cost (AC) is total cost (C) divided by the quantity (x). Marginal average cost is the derivative of AC. Pay close attention to distinguishing between total cost, average cost, marginal cost, and marginal average cost during calculations.

Question 7.
(i) Find the marginal revenue, if the average revenue is 45 and the elasticity of demand is 5.
(ii) Find the price, if the marginal revenue is 28 and elasticity of demand is 3.
(iii) Find the elasticity of demand, if the marginal revenue is 50 and price is Rs. 75.
Answer: Solution:
(i) Given R_A = 45 and \( \eta \) = 5 Now, \( R_m = R_A (1 - \frac{1}{\eta}) \) \( = 45(1-\frac{1}{5}) \) \( = 45(\frac{4}{5}) \) \( = 36 \) Hence, the marginal revenue = 36.
(ii) Given R_m = 28 and \( \eta \) = 3 Now, \( R_m = R_A(1-\frac{1}{\eta}) \)
\( \therefore 28=R_A(1-\frac{1}{3}) = \frac{2}{3}R_A \)
\( \therefore R_A = \frac{28 \times 3}{2} = 42 \) Hence, the price = 42.
(iii) Given R_m = 50 and R_A = 75 Now, \( R_m = R_A (1 - \frac{1}{\eta}) \)
\( \therefore 50 = 75(1 - \frac{1}{\eta}) \)
\( \therefore 1 - \frac{1}{\eta} = \frac{50}{75} = \frac{2}{3} \)
\( \therefore \frac{1}{\eta} = 1 - \frac{2}{3} = \frac{1}{3} \)
\( \therefore \eta = 3 \) Hence, the elasticity of demand = 3.
In simple words: This question applies the relationship between marginal revenue (R_m), average revenue (R_A), and elasticity of demand (\( \eta \)). Marginal revenue is calculated as R_A times (1 - 1/\( \eta \)). By rearranging this formula, we can find any of the three variables if the other two are known.

🎯 Exam Tip: Memorize the key formula: \( R_m = R_A (1 - \frac{1}{\eta}) \). This formula is crucial for solving problems involving marginal revenue, average revenue, and elasticity of demand. Be careful with algebraic rearrangements.

Question 8. If the demand function is D = \( \frac{p+6}{p-3} \), find the elasticity of demand at p = 4.
Answer: Solution: The demand function is \( D=\frac{p+6}{p-3} \)
\( \therefore \frac{dD}{dp} = \frac{d}{dp} (\frac{p+6}{p-3}) \)
\( = \frac{(p-3)\frac{d}{dp}(p+6) - (p+6)\frac{d}{dp}(p-3)}{(p-3)^2} \)
\( = \frac{(p-3)(1+0) - (p+6)(1-0)}{(p-3)^2} \)
\( = \frac{p-3-p-6}{(p-3)^2} = \frac{-9}{(p-3)^2} \) Elasticity of demand is given by \( \eta = \frac{-p}{D} \frac{dD}{dp} \)
\( = \frac{-p}{\frac{p+6}{p-3}} \times \frac{-9}{(p-3)^2} \)
\( = \frac{9p}{(p+6)(p-3)} \) When p = 4, then
\( \eta = \frac{9(4)}{(4+6)(4-3)} = \frac{36}{10 \times 1} = 3.6 \). Hence, the elasticity of demand at p = 4 is 3.6
In simple words: To find elasticity of demand, first calculate the derivative of the demand function (dD/dp) using the quotient rule. Then, use the elasticity formula \( \eta = \frac{-p}{D} \frac{dD}{dp} \) and substitute the given price p=4 into the resulting expression.

🎯 Exam Tip: The elasticity of demand formula \( \eta = \frac{-p}{D} \frac{dD}{dp} \) is essential. Remember to correctly apply the quotient rule for differentiation and substitute values accurately to avoid calculation errors.

Question 9. Find the price for the demand function D = \( \frac{2p+3}{3p-1} \), when elasticity of demand is \( \frac{11}{14} \).
Answer: Solution: The demand function is \( D=\frac{2p+3}{3p-1} \)
\( \therefore \frac{dD}{dp} = \frac{d}{dp} (\frac{2p+3}{3p-1}) \)
\( = \frac{(3p-1)\frac{d}{dp}(2p+3) - (2p+3)\frac{d}{dp}(3p-1)}{(3p-1)^2} \)
\( = \frac{(3p-1)(2 \times 1+0) - (2p+3)(3 \times 1-0)}{(3p-1)^2} \)
\( = \frac{6p-2-6p-9}{(3p-1)^2} = \frac{-11}{(3p-1)^2} \) Elasticity of demand is given by \( \eta = \frac{-p}{D} \frac{dD}{dp} \)
\( = \frac{-p}{\frac{2p+3}{3p-1}} \times \frac{-11}{(3p-1)^2} \)
\( = \frac{11p}{(2p+3)(3p-1)} = \frac{11p}{6p^2+7p-3} \) If \( \eta = \frac{11}{14} \), then
\( \frac{11}{14} = \frac{11p}{6p^2+7p-3} \)
\( \therefore 66p^2+77p-33 = 154p \)
\( \therefore 66p^2-77p-33 = 0 \)
\( \therefore 6p^2-7p-3 = 0 \)
\( \therefore (2p-3)(3p+1) = 0 \)
\( \therefore 2p-3=0 \)
\( \therefore p=\frac{3}{2} \) [ \( \because \) p>0]
In simple words: First, calculate dD/dp using the quotient rule. Substitute this and the demand function into the elasticity formula, \( \eta = \frac{-p}{D} \frac{dD}{dp} \). Equate the resulting expression to the given elasticity (\( \frac{11}{14} \)) and solve the quadratic equation for p, choosing the positive value for price.

🎯 Exam Tip: Be meticulous with algebraic simplification after applying the elasticity formula. Solving the resulting quadratic equation is often a critical step, and ensure you consider only economically relevant (e.g., positive) solutions for price.

Question 10. If the demand function is D = 50 - 3p - p² elasticity of demand at
(i) p = 5
(ii) p = 2. Comment on the result.
Answer: Solution: The demand function is D = 50 - 3p - p²
\( \therefore \frac{dD}{dp} = \frac{d}{dp} (50 - 3p - p^2) \)
\( = 0-3 \times 1-2p \)
\( = -3 - 2p \) Elasticity of demand is given by \( \eta = \frac{-p}{D} \frac{dD}{dp} \)
\( = \frac{-p}{(50-3p-p^2)} \times (-3-2p) \)
\( = \frac{p(3+2p)}{50-3p-p^2} \)
(i) When p = 5, then
\( \eta = \frac{5(3 + 2 \times 5)}{50-3(5)-(5)^2} = \frac{5 \times 13}{50-15-25} = \frac{65}{10} = 6.5 \). Since, \( \eta > 1 \), the demand is elastic.
(ii) When p = 2, then
\( \eta = \frac{2(3 + 2 \times 2)}{50-3(2)-(2)^2} = \frac{2 \times 7}{50-6-4} = \frac{14}{40} = \frac{7}{20} \). Since, 0 < \( \eta \) < 1, the demand is inelastic.
In simple words: First, find the derivative of the demand function. Then, use the elasticity formula to derive an expression for \( \eta \). Substitute p=5 and p=2 into this expression to find the elasticity at those prices, and comment on whether demand is elastic ( \( \eta \) > 1) or inelastic (0 < \( \eta \) < 1).

🎯 Exam Tip: When commenting on elasticity, remember these key interpretations: \( \eta \) > 1 implies elastic demand, \( \eta \) < 1 implies inelastic demand, and \( \eta \) = 1 implies unitary elastic demand. Showing the calculation for \( \eta \) for each price point is crucial.

Question 11. For the demand function D = \( 100 - \frac{p^2}{2} \), find the elasticity of demand at
(i) p = 10
(ii) p = 6 and comment on the results.
Answer: Solution: The demand function is \( D = 100 - \frac{p^2}{2} \)
\( \therefore \frac{dD}{dp} = \frac{d}{dp} (100 - \frac{p^2}{2}) \)
\( = 0 - \frac{1}{2} \times 2p = -p \) The elasticity of demand is given by \( \eta = \frac{-p}{D} \frac{dD}{dp} \)
\( = \frac{-p}{(100 - \frac{p^2}{2})} \times (-p) = \frac{p^2}{(100 - \frac{p^2}{2})} \)
(i) When p = 10, then
\( \eta = \frac{(10)^2}{(100 - \frac{(10)^2}{2})} = \frac{100}{100-50} = \frac{100}{50} = 2 \). Since, \( \eta > 1 \), the demand is elastic.
(ii) When p = 6, then
\( \eta = \frac{(6)^2}{(100 - \frac{(6)^2}{2})} = \frac{36}{100-18} = \frac{36}{82} = \frac{18}{41} \). Since, 0 < \( \eta \) < 1, the demand is inelastic.
In simple words: Calculate the derivative of the demand function (dD/dp) as -p. Substitute this into the elasticity formula to get an expression for \( \eta \). Then, calculate \( \eta \) for p=10 and p=6, noting whether the demand is elastic ( \( \eta \) > 1) or inelastic (0 < \( \eta \) < 1).

🎯 Exam Tip: Always state the elasticity formula and substitute the values clearly. The final comment on whether demand is elastic or inelastic based on the calculated \( \eta \) value is a required part of the answer.

Question 12. A manufacturing company produces, x items at a total cost of Rs.(40 + 2x). Their price is given as p = 120 - x. Find the value of x for which
(i) revenue is increasing
(ii) profit is increasing
(iii) Also find an elasticity of demand for price 80.
Answer: Solution:
(i) The total revenue R is given by R = p.x = (120 - x)x
\( \therefore R = 120x - x^2 \)
\( \therefore \frac{dR}{dx} = \frac{d}{dx} (120x-x^2) \)
\( = 120 \times 1 - 2x \)
\( = 120 - 2x \) If the revenue is increasing, then \( \frac{dR}{dx} > 0 \)
\( \therefore 120 - 2x > 0 \)
\( \therefore 120 > 2x \)
\( \therefore x < 60 \) Hence, the revenue is increasing when x < 60.
(ii) Profit \( \pi = R-C \)
\( = (120x - x^2) - (40 + 2x) \)
\( = 120x - x^2 - 40 - 2x \)
\( = 118x - x^2 - 40 \)
\( \therefore \frac{d\pi}{dx} = \frac{d}{dx} (118x-x^2 - 40) \)
\( = 118 \times 1 - 2x - 0 \)
\( = 118 - 2x \) If the profit is increasing, then \( \frac{d\pi}{dx} > 0 \)
\( \therefore 118 - 2x > 0 \)
\( \therefore 118 > 2x \)
\( \therefore x < 59 \) Hence, the profit is increasing when x < 59.
(iii) \( p = 120 - x \)
\( \therefore x = 120 - p \)
\( \therefore \frac{dx}{dp} = \frac{d}{dp} (120-p) \)
\( = 0-1 \)
\( = -1 \) Elasticity of demand is given by \( \eta = \frac{-p}{x} \frac{dx}{dp} \)
\( = \frac{-p}{(120-p)} \times (-1) = \frac{p}{120-p} \) When p = 80, then
\( \eta = \frac{80}{(120-80)} = \frac{80}{40} = 2 \).
In simple words: For revenue to increase, find the derivative of the total revenue (R=P.x) and set it > 0. For profit to increase, find the derivative of profit (π=R-C) and set it > 0. For elasticity, rearrange the demand function to get x in terms of p, then use the elasticity formula \( \eta = \frac{-p}{x} \frac{dx}{dp} \) at p=80.

🎯 Exam Tip: This question combines several concepts. Clearly label each part (i), (ii), (iii). When finding elasticity, if the demand function is given as p in terms of x, it's often easier to find dx/dp rather than dD/dp, then use \( \eta = \frac{-p}{x} \frac{dx}{dp} \).

Question 13. Find MPC, MPS, APC and APS, if the expenditure E_c of a person with income I is given as E_c = (0.0003)I² + (0.075)I, when I = 1000.
Answer: Solution: E_c = (0.0003)I² + (0.075)I
\( MPC = \frac{dE_c}{dI} = \frac{d}{dI} [(0.0003)I^2 + (0.075)I] \)
\( = (0.0003)(2I) + (0.075)(1) \)
\( = (0.0006)I + 0.075 \) When I = 1000, then MPC = (0.0006)(1000) + 0.075 \( = 0.6 + 0.075 \) \( = 0.675 \).
\( \therefore MPC + MPS = 1 \)
\( \therefore 0.675 + MPS = 1 \)
\( \therefore MPS = 1 - 0.675 = 0.325 \) Now, \( APC = \frac{E_c}{I} = \frac{(0.0003)I^2+(0.075)I}{I} \)
\( = (0.0003)I + (0.075) \) When I = 1000, then APC = (0.0003)(1000) + 0.075 \( = 0.3 + 0.075 \) \( = 0.375 \)
\( \therefore APC + APS = 1 \)
\( \therefore 0.375 + APS = 1 \)
\( \therefore APS = 1 - 0.375 = 0.625 \) Hence, MPC = 0.675, MPS = 0.325, APC = 0.375, APS = 0.625.
In simple words: To find MPC (Marginal Propensity to Consume), differentiate the expenditure function with respect to income (I). To find APC (Average Propensity to Consume), divide the expenditure function by income (I). Then, use the relationships MPC + MPS = 1 and APC + APS = 1 to find MPS (Marginal Propensity to Save) and APS (Average Propensity to Save).

🎯 Exam Tip: Understand the definitions: MPC is dE_c/dI, APC is E_c/I. Remember the fundamental identities: MPC + MPS = 1 and APC + APS = 1. Accuracy in differentiation and substitution is key for these calculations.