Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.4 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.4

Question 1.

The demand function of a commodity at price P is given as D = 40 – \(\frac{5 P}{8}\). Check whether it is an increasing or decreasing function.

Solution:

D = 40 – \(\frac{5 P}{8}\)

∴ \(\frac{d D}{d P}=\frac{d}{d P}\left(40-\frac{5 P}{8}\right)\)

= 0 – \(\frac{5}{8}\) × 1

= \(\frac{-5}{8}\)

Hence, the given function is decreasing function.

Question 2.

The price P for demand D is given as P = 183 + 120D – 3D^{2}, find D for which price is increasing.

Solution:

P = 183 + 120D – 3D^{2}

∴ \(\frac{d P}{d D}=\frac{d}{d D}\)(183 + 120D – 3D^{2})

= 0 + 120 × 1 – 3 × 2D

= 120 – 6D

If price P is increasing, then \(\frac{d P}{d D}\) > 0

∴ 120 – 6D > 0

∴ 120 > 6D

∴ D < 20

Hence, the price is increasing when D < 20.

Question 3.

The total cost function for production of x articles is given as C = 100 + 600x – 3x^{2}. Find the values of x for which the total cost is decreasing.

Solution:

The cost function is given as

C = 100 + 600x – 3x^{2}

∴ \(\frac{d C}{d D}=\frac{d}{d D}\)(100 + 600x – 3x^{2})

= 0 + 600 × 1 – 3 × 2x

= 600 – 6x

If the total cost is decreasing, then \(\frac{d C}{d x}\) < 0

∴ 600 – 6x < 0

∴ 600 < 6x

∴ x > 100

Hence, the total cost is decreasing for x > 100.

Question 4.

The manufacturing company produces x items at the total cost of ₹(180 + 4x). The demand function for this product is P = (240 – x). Find x for which

(i) revenue is increasing

(ii) profit is increasing.

Solution:

(i) Let R be the total revenue.

Then R = P.x = (240 – x)x

∴ R = 240x – x^{2}

∴ \(\frac{d R}{d D}=\frac{d}{d D}\)(240x – x^{2})

= 240 × 1 – 2x

= 240 – 2x

R is increasing, if \(\frac{d R}{d x}\) > 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 π = R – C

∴ π = (240x – x^{2}) – (180 + 4x)

= 240x – x^{2} – 180 – 4x

= 236x – x^{2} – 180

∴ \(\frac{d \pi}{d x}=\frac{d}{d x}\)(236x – x^{2} – 180)

= 236 × 1 – 2x – 0

= 236 – 2x

Profit is increasing, if \(\frac{d \pi}{d x}\) > 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.

Question 5.

For manufacturing x units, labour cost is 150 – 54x and processing cost is x^{2}. Price of each unit is p = 10800 – 4x^{2}. Find the values of x for which

(i) total cost is decreasing

(ii) revenue is increasing.

Solution:

(i) Total cost C = labour cost + processing cost

∴ C = 150 – 54x + x^{2}

∴ \(\frac{d C}{d x}=\frac{d}{d x}\)(150 – 54x + x^{2})

= 0 – 54 × 1 + 2x

= -54 + 2x

The total cost is decreasing, if \(\frac{d C}{d x}\) < 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}

∴ \(\frac{d R}{d x}=\frac{d}{d x}\)(10800x – 4x^{3})

= 10800 × 1 – 4 × 3x^{2}

= 10800 – 12x^{2}

The revenue is increasing, if \(\frac{d R}{d x}\) > 0

i.e. if 10800 – 12x^{2} > 0

i.e. if 10800 > 12x^{2}

i.e. if x^{2} < 900

i.e. if x < 30 ……[∵ x > 0]

Hence, the revenue is increasing, if x < 30.

Question 6.

The total cost of manufacturing x articles is C = 47x + 300x^{2} – x^{4}. Find x, for which average cost is

(i) increasing

(ii) decreasing.

Solution:

The total cost is given as C = 47x + 300x^{2} – x^{4}

∴ the average cost is given by

(i) CA is increasing, if \(\frac{d C_{A}}{d x}\) > 0

i.e. if 300 – 3x^{2} > 0

i.e. if 300 > 3x^{2}

i.e. if x^{2} < 100

i.e. if x < 10 …..[∵ x > 0]

Hence, the average cost is increasing, if x < 10.

(ii) CA is decreasing, if \(\frac{d C_{A}}{d x}\) < 0

i.e. if 300 – 3x^{2} < 0

i.e. if 300 < 3x^{2
}i.e. if x^{2} > 100

i.e. if x > 10 ……[∵ x > 0]

Hence, the average cost is decreasing, if x > 10.

Question 7.

(i) Find the marginal revenue, if the average revenue is 45 and the elasticity of demand is 5.

Solution:

Given R_{A} = 45 and η = 5

Now, R_{m} = \(R_{A}\left(1-\frac{1}{\eta}\right)\)

= 45(1 – \(\frac{1}{5}\))

= 45(\(\frac{4}{5}\))

= 36

Hence, the marginal revenue = 36.

(ii) Find the price, if the marginal revenue is 28 and elasticity of demand is 3.

Solution:

Given R_{m} = 28 and η = 3

Hence, the price = 42.

(iii) Find the elasticity of demand, if the marginal revenue is 50 and price is ₹ 75.

Solution:

Given R_{m} = 50 and R_{A} = 75

Hence, the elasticity of demand = 3.

Question 8.

If the demand function is D = \(\frac{p+6}{p-3}\), find the elasticity of demand at p = 4.

Solution:

The demand function is

Elasticity of demand is given by

Hence, the elasticity of demand at p = 4 is 3.6

Question 9.

Find the price for the demand function D = \(\frac{2 p+3}{3 p-1}\), when elasticity of demand is \(\frac{11}{14}\).

Solution:

The demand function is

Elasticity of demand is given by

Question 10.

If the demand function is D = 50 – 3p – p^{2} elasticity of demand at (i) p = 5 (ii) p = 2. Comment on the result.

Solution:

The demand function is D = 50 – 3p – p^{2}

∴ \(\frac{d D}{d p}=\frac{d}{d p}\)(50 – 3p – p^{2})

= 0 – 3 × 1 – 2p

= -3 – 2p

Elasticity of demand is given by

(i) When p = 5, then

Since, η >1, the demand is elastic.

(ii) When p = 2, then

Since, 0 < η < 1, the demand is inelastic.

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.

Solution:

The demand function is

The elasticity of demand is given by

(i) When p = 10, then

Since, η > 1, the demand is elastic.

(ii) When p = 6, then

Since, 0 < η < 1, the demand is inelastic.

Question 12.

A manufacturing company produces, x items at a total cost of ₹(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.

Solution:

(i) The total revenue R is given by

R = p.x = (120 – x)x

∴ R = 120x – x^{2}

∴ \(\frac{d R}{d x}=\frac{d}{d x}\)(120x – x^{2})

= 120 × 1 – 2x

= 120 – 2x

If the revenue is increasing, then \(\frac{d R}{d x}\) > 0

∴ 120 – 2x > 0

∴ 120 > 2x

∴ x < 60

Hence, the revenue is increasing when x < 60.

(ii) Profit π = R – C

= (120x – x^{2}) – (40 + 2x)

= 120x – x^{2} – 40 – 2x

= 118x – x^{2} – 40

∴ \(\frac{d \pi}{d x}=\frac{d}{d x}\)(118x – x^{2} – 40)

= 118 × 1 – 2x – 0

= 118 – 2x

If the profit is increasing, then \(\frac{d \pi}{d x}\) > 0

∴ 118 – 2x > 0

∴ 118 > 2x

∴ x < 59

Hence, the profit is increasing when x < 59.

(iii) p = 120 – x

∴ x = 120 – p

∴ \(\frac{d x}{d p}=\frac{d}{d p}\)(120 – p)

= 0 – 1

= -1

Elasticity of demand is given by

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)I^{2} + (0.075)I, when I = 1000.

Solution:

E_{c} = (0.0003)I^{2} + (0.075)I

MPC = \(\frac{d E_{c}}{d I}=\frac{d}{d I}\)[(0.0003)I2 + (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.

∴ MPC + MPS = 1

∴ 0.675 + MPS = 1

∴ 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

∵ APC + APS = 1

∴ 0.375 + APS = 1

∴ APS = 1 – 0.375 = 0.625

Hence, MPC = 0.675, MPS = 0.325, APC = 0.375, APS = 0.625.