Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 3 Integral Calculus II Miscellaneous Problems Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Miscellaneous Problems

Question 1.

A manufacture’s marginal revenue functional is given by MR = 275 – x – 0.3x². Find the increase in the manufactures total revenue if the production increased from 10 to 20 units.

Solution:

MR = 275 – x – 0.3x²

The increase in the manufactures total revenue 20

T.R = ∫MR dx = (275 – x – 0.3x²) dx

= [5500 – 200 – 0.1 (8000)] – [2750 – 50 – 0.1(1000)]

= [5500 – 200 – 800] – [2750 – 50 – 100]

= 4500 – 2600

= Rs 1900

Question 2.

A company has determined that marginal cost function for product of a particular commodity is given by MC = 125 + 10x – \(\frac { 8 }{3}\). Where C is the cost of producing x units of the commodity. If the fixed cost Rs 250 what is cost of producing 15 units.

Solution:

MC = 125 + 10x – \(\frac { x^2 }{9}\)

Fixed cos t K = Rs 250

C = ∫MC dx – ∫(125 + 10x – \(\frac { x^2 }{9}\)) dx

C = 125x + \(\frac { 10x^2 }{9}\) – \(\frac { x^3 }{9×3}\) + k

C = 125x + 5x² – \(\frac { x^3 }{27}\) + 250

when x = 15

C = 125(15) + 5(15)² – \(\frac { (15)^3 }{27}\) + 250

= 1875 + 1125 – 125 + 250

C = Rs 3,125

Question 3.

The marginal revenue function for a firm given by MR = \(\frac { 2 }{x+3}\) – \(\frac { 2x }{(x+3)^2}\) + 5. Show that the demand function is P = \(\frac { 2x }{(x+3)^2}\) + 5

Solution:

Question 4.

For the marginal revenue function MR = 6 – 3x² – x³, Find the revenue function and demand function

Solution:

MR = 6 – 3x² – x³

Revenue function R = ∫MR dx

R = ∫(6 – 3x² – x³)dx

Question 5.

The marginal cost of production of a firm is given by C'(x) = 20 + \(\frac { x }{20}\) the marginal revenue is given by R’(x) = 30 and the fixed cost is Rs 100. Find the profit function.

Solution:

C'(x) = 20 + \(\frac { x }{20}\)

Fixed cost k_{1} = Rs 100

C(x) = ∫C_{1}(x) dx + k_{1}

Question 6.

The demand equation for a product is P_{d} = 20 – 5x and the supply equation is P_{s} = 4x + 8. Determine the consumers surplus and producer’s surplus under market equilibrium.

Solution:

P_{d} = 20 – 5x and P_{s} = 4x + 8

At market equilibrium

P_{d} = P_{d}

20 – 5x = 4x + 8 ⇒ 20 – 8 = 4x + 5x

9x = 12 ⇒ x = \(\frac { 12 }{9}\)

∴ x = \(\frac { 4 }{3}\)

Question 7.

A company requires f(x) number of hours to produce 500 units. lt is represented by f(x) = 1800x^{-0.4}. Find out the number of hours required to produce additional 400 units. [(900)^{0.6} = 59.22, (500)^{0.6} = 41.63]

Solution:

f(x) number of hours to produce 500

f(x) = 1800 x^{-0.4}

The number of hours required to produce additions

Question 8.

The price elasticity of demand for a commodity is \(\frac { p }{x^3}\). Find the demand function if the quantity of demand is 3, When the price is Rs 2.

Solution:

Price elasticity of demand

Question 9.

Find the area of the region bounded by the curve between the parabola y = 8x² – 4x + 6 the y-axis and the ordinate at x = 2.

Solution:

Equation of the parabola

y = 8x² – 4x + 6

The required region is bounded by the y-axis and the ordinate at x = 2.

∴ Required Area A = \(\int_{0}^{2}\)ydx

Question 10.

Find the area of the region bounded by the curve y² = 27x³ and the line x = 0, y = 1 and y = 2

Solution:

Equation of the curve is y² = 27x³

⇒ x³ = \(\frac { y^2 }{27}\) = \(\frac { y^3 }{3^3}\)

∴ x = \(\frac { (y)^{2/3} }{3}\)

Since the Area of the region bounded by the given curve and the lines x = 0, y = 1 and y = 2

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