Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 3 Integral Calculus II Ex 3.3 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 3 Integral Calculus II Ex 3.3

Question 1.

Calculate consumer’s surplus if the demand function p = 50 – 2x and x = 20

Solution:

Demand function p = 50 – 2x and x = 20

when x = 20, p = 50 – 2(20)

p = 50 – 40 = 10

∴ p_{0} = 10

CS = \(\int _{0}^{x}\) (demand function) dx – (Price × quantity demanded)

= \(\int _{0}^{20}\) (50 – 2x)dx – (10 × 20)

= [50x – 2(\(\frac { x^2 }{2}\))]\( _{0}^{20}\) – 200

= [50x – x²]\( _{0}^{20}\) – 200

= {50(20) – (20)² – [0]} – 200

= (1000 – 400) – 200

= 600 – 200

∴ C.S = 400 units

Question 2.

Calculate consumer’s surplus if the demand function p = 122 – 5x – 2x² and x = 6.

Solution:

Demand function p = 122 – 5x – 2x² and x = 6

when x = 6; p = 122 – 5(6) – 2(6)²

= 122 – 30 – 2 (36)

= 122 – 102 = 20

∴ p_{0} = 20

C.S = \(\int _{0}^{x}\) (demand function) dx – (Price × quantity demanded)

= \(\int _{0}^{6}\)(122 – 5x – 2x²) dx – (20 × 6)

[732 – 5(18) – 2(72)] – 120

= 732 – 90 – 144 – 120

= 732 – 354 = 378

∴ CS = 378 units

Question 3.

The demand function p = 85 – 5x and supply function p = 3x – 3. Calculate the equilibrium price and quantity demanded. Also calculate consumer’s surplus.

Solution:

Demand function p = 85 – 5x

Supply function p = 3x – 35

W.K.T. at equilibrium prices p_{d} = p_{s}

85 – 5x = 3x – 35

85 + 35 = 3x + 5x

120 = 8x ⇒ x = \(\frac { 120 }{8}\)

∴ x = 15

when x = 15 p_{0} = 85 – 5(15) = 85 – 75 = 10

C.S = \(\int _{0}^{x}\) f(x) dx – x_{0}p_{0}

= \(\int _{0}^{x}\) (85 – 5x) dx – (15)(10)

= 1275 – \(\frac { 1125 }{2}\) – 150

= 1275 – 562.50 – 150

= 1275 – 712.50

∴ CS = 562.50 units

Question 4.

The demand function for a commodity is p = e^{-x}. Find the consumer’s surplus when p = 0.5

Solution:

The demand function p = e^{-x}

when p = 0.5 ⇒ 0.5 = e^{-x}

\(\frac { 1 }{2}\) = \(\frac { 1 }{e^x}\) ⇒ e^{x} = 2

∴ x = log 2

∴ Consumer’s surplus

C.S = \(\int _{0}^{x}\) (demand function) dx – (Price × quantity demanded)

Question 5.

Calculate the producer’s surplus at x = 5 for the supply function p = 7 + x

Solution:

The supply function p = 7 + x

when x = 5 ⇒ p = 7 + 5 = 12

∴ x_{0} = 5 and p_{0} = 12

Producer’s surplus

Question 6.

If the supply function for a product is p = 3x + 5x². Find the producer’s surplus when x = 4

Solution:

The supply function p = 3x + 5x²

when x = 4 ⇒ p = 3(4) + 5(4)²

p = 12 + 5(16)= 12 + 80

p = 92

∴ x_{0} = 4 and p_{0} = 92

Producer’s Surplus

Question 7.

The demand function for a commodity is p = \(\frac { 36 }{x+4}\) Find the producer’s surplus when the prevailing market price is Rs 6.

Solution:

The demand function for a commodity

p = \(\frac { 36 }{x+4}\)

when p = 6 ⇒ 6 = \(\frac { 36 }{x+4}\)

x + 4 = \(\frac { 36 }{6}\) ⇒ x + 4 = 6

x = 2

∴ p_{0} = 6 and x_{0} = 2

The consumer’s surplus

C.S = \(\int _{0}^{x}\) f(x) dx – x_{0}p_{0}

= \(\int _{0}^{2}\) (\(\frac { 36 }{x+4}\)) dx – 2(6)

= 36 [log (x + 4)]\(_{0}^{2}\) – 12

= 36 [log (2 + 4) – log (0 + 4)] – 12

= 36 [log6 – log4] – 12

= 36[log(\(\frac { 6 }{4}\))] – 12

∴ CS = 36 log(\(\frac { 6 }{4}\)) – 12 units

Question 8.

The demand and supply functions under perfect competition are pd = 1600 – x² and p_{s} = 2x² + 400 respectively, find the producer’s surplus.

Solution:

pd = 1600 – x² and p_{s} = 2x² + 400

Under the perfect competition p_{d} = p_{s}

1600 – x² = 2x² + 400

1600 – 400 = 2x² + x² ⇒ 1200 = 3x²

⇒ x² – 400 ⇒ x = 20 or -20

The value of x cannot be negative, x = 20 when x_{0} = 20;

p_{0} = 1600 – (20)² = 1600 – 400

P_{0} = 1200

PS = x_{0}p_{0} – \(\int _{0}^{x_0}\) g(x) dx

Question 9.

Under perfect competition for a commodity the demand and supply laws are P_{d} = \(\frac { 8 }{x+1}\) – 2 and Ps = \(\frac { x+3 }{2}\) respectively. Find the consumer’s and producer’s surplus.

Solution:

16 – (x² + 3x + x + 3) = 2 [2(x + 1)]

16 – (x² + 4x + 3) = 4(x + 1)

16 – x² – 4x – 3 = 4x + 4

x² + 4x + 4x + 4 + 3 – 16 = 0

x² + 8x – 9 = 0

(x + 9) (x – 1) = 0 ⇒ x = -9 (or) x = 1

The value of x cannot be negative x = 1 when x_{0} = 1

p_{0} = \(\frac { 8 }{1+1}\) – 2 ⇒ p_{0} = \(\frac { 8 }{2}\) – 2

p_{0} = 4 – 2 ⇒ p_{0} = 2

CS = \(\int _{0}^{x}\) f(x) dx – x_{0}p_{0}

= \(\int _{0}^{1}\) (\(\frac { 8 }{x+1}\) – 2) dx – (1) (2)

= {8[log(x + 1)] – 2x}\(\int _{0}^{1}\) – 2

= 8 {[log (1 + 1) – 2(1)] – 8 [log (0 + 1) – 2(0)]} – 2

= [8 log (2) – 2 – 8 log1] – 2

= 8 log(\(\frac { 8 }{2}\)) – 2 – 2

C.S = (8 log 2 – 4) units

P.S = x_{0}p_{0} – \(\int _{0}^{x_0}\) g(x) dx

Question 10.

The demand equation for a products is x = \(\sqrt {100-p}\) and the supply equation is x = \(\frac {p}{2}\) – 10. Determine the consumer’s surplus and producer’s, under market equilibrium.

Solution:

p_{d} = \(\sqrt {100-p}\) and p_{s} = \(\sqrt {100-p}\)

Under market equilibrium, p_{d} = p_{s}

\(\sqrt {100-p}\) = \(\frac {p}{2}\) – 10

Squaring on both sides

36p = p² ⇒ p² – 36 p = 0

p (p – 36) = 0 ⇒ p = 0 or p = 36

The value of p cannot be zero, ∴p_{0} = 36 when p_{0} = 36

x_{0} = \(\sqrt {100-36}\) = \(\sqrt {64}\)

∴ x_{0} = 8

= 288 – [x² + 20x]\( _{0}^{8}\)

= 288 – { [(8)² + 20(8)] – [0]}

= 288 – [64 + 160]

= 288 – 224 = 64

PS = 64 Units

Question 11.

Find the consumer’s surplus and producer’s surplus for the demand function pd = 25 – 3x and supply function ps = 5 + 2x

Solution:

p_{d} = 25 – 3x and p_{s} = 5 + 2x

Under market equilibrium, p_{d} = p_{s}

25 – 3x = 5 + 2x

25 – 5 = 2x + 3x ⇒ 5x = 20

∴ x = 4

when x = 4

P_{0} = 25 – 3(4)

= 25 – 12 = 13

p_{0} = 13

= 52 – [5x + x²]\( _{0}^{4}\)

= 52 – (5(4) + (4)²) – (0)}

= 52 – [20 + 16]

= 52 – 36

∴ PS = 16 units