An industrial firm makes two products, A and B. Water is needed to make each uni
ID: 329108 • Letter: A
Question
An industrial firm makes two products, A and B. Water is needed to make each unit of A and B. The production functions relating the amount of water XA needed to make A, and the amount of water XB needed to make B are A = 0.5 XA, and B = 0.25 XB,respectively. Water is the scarce resource-they have plenty of other needed resources. The products they make are unique, and hence they can set the unit price of each product at any value they want to. However experience tells them that the higher the unit price for a product, the less amount of that product they will sell. The relationship between unit price and quantity that can be sold is given by the following two demand functions. (a) What are the amounts of A and B, and their unit prices, that maximize the total revenue that can be obtained?
(b) Suppose the total amount of A and B could not exceed some amount Tmax. What are the amounts of A and B, and their unit prices, that maximize total revenue, if
i) Tmax = 10
ii) Tmax = 5
(c) Use linear programming to find the amounts of A and B and their unit prices that maximize total revenue assuming the total amount of water available is 10 units
Po Po 8-A Unit - 1.5B Unit price Price Quantity of product A Quantity of product BExplanation / Answer
(a) Let A and B be the prices of product A and B
Amount of A = 8-A,
Amount of B = 6-1.5B
Revenue from A, R(A)= (8-A)*A = 8A - A2
Revenue from B, R(B) = (6-1.5B)*B = 6B - 1.5B2
Taking the first order derivative of R(A), d(R(A)/dA = d(8A-A2)/dA = 8 - 2A. At maxima, slope is equal to o, so 8-2A = 0 or A = 4
Similarly, first order of R(B), d(R(B)/dB = d(6B-1.5B2)/dB = 6 - 3B. Equating this to 0, we get, B = 2
Therefore, revenue maximizing prices are
A = 4, quantity = 8-4 = 4
B = 2, quantity = 6-1.5*2 = 3
Total revenue = 4*4+2*3 = 22
(b) i) Tmax = 10
In this case, solution as determined in part (a) applies, as the total quantity of part a = 4+3 = 7, which is less than 10. So solution is same as part (a)
ii) Tmax = 5
Price of B can be either 1 or 2, otherwise, quantity will be either 0 or fraction. So we keep it 2, hence quantity of B = 6-1.5*2 = 3 .
Therefore, quantity of A can be 2 only. Hence, price of A = 8-2 = 6
Total revenue = 2*6+3*2 = 18