MAYJUNE 2017 EXAM SOLUTIONS AND QUESTION

1. 4 Point A is the intercept for the demand function since the demand function slopes downwards. Using the demand funct...

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1. 4 Point A is the intercept for the demand function since the demand function slopes downwards. Using the demand function when Q = 0 P = 952 + 8(0) = 952 Point B is the intercept for the supply function

P = 400 + 4(0) = 400 Point C use the demand function when P = 0 0 = 952 -8Q Q = 119

2. 2 Point D is the equilibrium point and it is calculated as dd function = ss function 952 – 8Q = 400 + 4Q Solve for Q Q = 46 Solve for P using the demand function P = 952 – 8(46) = 584

3. 1

= -1/5 When Q=8 P = 60

Price elasticity of demand = -1/5 x 60/8 = - 1.5

4. 4 At breakeven point Total revenue = total cost P = 100 – 20Q Total revenue equals price, P, times quantity, Q, or TR = P×Q Total revenue = Q(100 – 20Q)

Total cost = Total variable cost + fixed costs = 10Q + 40

At BEP Q(100 – 20Q) = 10Q +40 100Q -20Q^2 =10Q +40 20Q^2 -90Q +40 = 0 Solve for Q using quadratic formula and get 0.5 or 4

5. 1 = (-3)/5

Using this as the slope we can derive the equation as below (80 – Q)/(40 – P) = -3/5 80 –Q = -24 + 3/5P Making Q the subject and simplifying we get Q = 104 - 3/5P = 104 -0.6P

6. 3 Marginal Cost function is a derivative of the total cost function. So to get the total cost from the marginal cost we have to find the integral of the MC and the fixed cost will be the constant.

C = 100

7. 2

Factorize the numerator as shown below

8. 1

Profit = TR – TC = 40Q -8Q^2 –(8 +16Q-Q^2) = -7Q^2 + 24Q -8 Solve for the maximum point of the profit function

Q = 12/7 X 1000 = 1 714.2857

9. 3

You need to calculate area under the demand function bounded by 0 and 6 Calculate the area as follows;

10. 2 TC = Variable cost + Fixed cost = 17Q + 744 MC = Derivative of the TC or = Variable cost = 17

11. 4 First solve for the equilibrium Pd = Ps 136 -4Q =14 +5Q Solve and get Q = 13.56 P = 81.78

81.7 8

Area = 459.55

649.38

14 Area =189.84

13.56

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1

Solution is displayed as a decimal version of this answer.

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2

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Fibre 10x + 20y + 30z= 1800 X + 2y +3z = 180 Fat 30x+ 20y+40z= 2800 3x +2y + 4z = 280 Protein 20x + 40y + 25 z = 2200 4x+10y + 5z= 440

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The lowest point that minimizes z is (3.4)

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At equilibrium Pd = Ps 150 -6Q^2 = 10Q^2 +20Q 16Q^2 +20Q -150 = 0 Solve for Q = 3 and P = 96 (150-6(3^2))

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4

Roots of the equation = -3 and -1 as shown on the graph Equation will be as follows:

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Find the value of the derivative of the equation when t =15 The derivate is as follows

=8.59/15 =0.572667 x 1 000 000 = 572 667

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1

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When p = 6 q = 156

= 2p = 2(6) = 12

PED =-12 X 6/156 =- 0.461538

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First find the total revenue TR = Q X P = P(80 –P^2) Marginal revenue is the derivative of the TR curve Derivative = 80 -3p^2

Where p = 3 the Marginal Revenue = 80 -3(3^2) = 53

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