Subject Code: PQC1C

Paper: QUANTITATIVE TECHNIQUE (1C)

Specific Instructions:

Answer all the four questions. Marks allotted 100. Each Question carries equal marks.

General Instructions:

The Student should submit this assignment in the handwritten form (not in the typed format) The Student should submit this assignment within the time specified by the exam dept Each Question mentioned in this assignment should be answered within the word limit specified The student should only use the Rule sheet papers for answering the questions. The student should attach this assignment paper with the answered papers. Failure to comply with the above Five instructions would lead to rejection of assignment.

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Question No 1

( A )

Differentiate the following functions with respect to x:

a) y = e√ax + b b) y = log ax + b cx + d

c) y = ax + xa

a) (p)= d/dx e√ax + b = root ax + B

b) p= D log( ax+b/ Cx+d) /dy

= 1/ a/c = C/a

c) xax-1 + a

P=d a^x + X^a/ dy

= A x ^x-1 + aXa^a-1

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(B)

(a) If y = √x + (1/√x) , show that 2x (dy / dx) – y + (2 /√x) = 0

Answer

y = [x - {1- (1/x) }^-1 ]^-2 ==> 1/ [x - { (x - 1)/x) }^-1 ]^2

==> 1/ [x - { x /(x - 1) } ]^2 ==> 1/ [(x^2 - x - x) /(x - 1) ]^2

==> 1/ [(x^2 - 2x ) /(x - 1) ]^2 ==> (x - 1) ^2 / [(x^2 - 2x )^2

Now to differentiate this use u/v rule i.e. (d/dx)(u/v) = [v*u' - u*v']/v^2

so dy/dx = [(x^2-2x)^2*2(x-1) - (x -1)^2*2(x^2-2x)*(2x-2)]/(x^2 - 2x )^4

=> 2(x-1)(x^2-2x) [(x^2-2x) - 2 (x - 1)^2]/(x^2 - 2x )^4

=> 2 [(x^2-2x) - 2 (x - 1)^2]/(x^2 - 2x )^3

=> -2 [(x^2 - 2x + 2 ]/(x^2 - 2x )^3

b) If y = 2x/(x + 6) , prove that 2x (dy / dx) = y (2- y)

Answer

Comparing the given differential equation with the general first order differential equation, we have

P(x) = -2 x and Q(x) = x

Let us now find the integrating factor u(x)

u(x) = e P(x) dx

= e -2 x dx

= e - x2

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We now substitute u(x)= e - x2 and Q(x) = x in the equation u(x) y = u(x) Q(x) dx to obtain

e - x2 y = x e - x2dx

Integrate the right hand term to obtain

e - x2 y = -(1/2) e-x2 + C , C is a constant of integration.

Solve the above for y to obtain

y = C ex2 - 1/2

As a practice, find dy / dx and sustitute y and dy / dx in the given equation to check that the solution found is correct.

(C ) Evaluate the first and second order partial derivatives of

u = (x2 + 2xy – y2).ex

and show ∂2u/∂x∂y = ∂2u/∂y∂x.

Answer

(D )The demand and supply relations are given by the equations:

p2 + q2 =20 and 2p + q = 8

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where p is the price and q is the quantity. Determine the equilibrium price and

quantity.

Answer

p2 + (8- 2p) ^2 =20

after solving this equation

P = 2 and q = -4

So the equilibrium price would be the same.

Question No 2

A travel agency has chartered a bus to Shimla for college students. The charge is Rs.100 with an additional charge of Rs.2.50 for each subsequent cancellation. The bus has a seating capacity of 50 passengers. Determine the total revenue R(x), as a function of x, the number of cancellations received prior to the departure date. What is the value of x for which R(x) is maximum?

P(x) = - 50x2 +100x + 2.50

P'(x) = -2x + 100 = 0

x = 50

Question No 3

(i) The production team of Revlon calculates that it can sell out all the products that it produces at the rate of Rs.2 per unit. It estimates the cost function for the product to be: 1000+0.5 (Q/50)2 for Q units produced. Find the number of units sold that will maximise the company’s profit; and what is the amount of the maximum profit?

Answer

P (X) = 1000 + ½ (q^2/2500)

P`(X) = 100-.5Q = MC = 2Q1000 – 0.5Q = 2Q

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100 0= 2Q + 0.5Q100 0= 2.5Q100/02.5=Q400=Q*

Hence producing Q*=400 will maximise profits, as the monopolist captures allof the consumer surplus

(ii) The function describing Marginal Cost is defined as 4x +1,500, where x is the number of units produced. It is estimated that the total cost equals Rs.40,200 for every 10 units produced. Determine the total cost function.

Answer

C(X) = Intg (4x +1,500) dx

= 4x^2/2+ 500x+ K

K is a constant and the c (O) = FC= 10

So the cost function would be

C (X) = 4x^2/2+ 500x+4020

Question No 4

( A) Verify the relationship MR = p[1 – 1/ηd] for the demand functions: (i) p = 86 – 25x (ii) p = 280/x2

(I) P = 86-25x

TR→MAX => MR=0

86 = 25xX = 86/25 = 3.44

(ii) p = 280/x2

MR=(TR)'= 560xTR=4.62475*63.002 = 291.369TC=265.446Profit=TR-TC=25.9223

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( B ) Given the marginal cost function of a firm MC = 2 – 4x + 3x2 , find total variable cost when output x is : (a) 4 and (b) 10

(I) MC = 2 – 4x + 3x

4x-2=3x+10-3x -3x

1x-2=10+2 +2

1x=20

Since the two X's are on opposite sides you have to subtract 3x from itself and 4x. The 3x 's cancel out. That gives you 1x-2=10. Than you add the 2 to the 2 and to 18, the 2's cancel out and leaves you with 20. Than you need to get rid of the x so you divide 1x ino 20 and get 20.

(II)MC = 2 – 4x + 3x2

MC(q) = 0.012 q2 - 0.14 q + 5.4

Total variable cost = .12*4 +.14*4+5.4 = .4.8 + .56 +5.4 = 8

( C ) When the price of an umbrella is p rupees, x umbrellas are demanded, and when the price of a raincoat is q rupees, y raincoats are demanded. The respective demand equations are:

x = 4e-p/100q and y = 8e-q/200p

Mathematically establish what kind of market relationship exists between the two products.

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