Exercises - Math Problem Example

Assignment 4, Due July 26th in A firm manufactures and sells q units of a product at price = (575 - q) which has unit costs of (q2 -25q) and fixed costs of 45,000. (a) Write down expressions for: revenue, profit and average cost in terms of output(q) of the firm. [1 mark] Revenue = (575 - q ) q = 575q - q2 Profit = Revenue - Total Cost = 575q - q2 - [(q2 - 25q )q +45,000] = 575q - q2 - q3 + 25q2 - 45,000 = - q3 + 24.5q2 + 575q - 45,000 Average Cost = Total Cost / q = q2 - 25q + 45,000/q (b) Find expressions for: marginal revenue, marginal cost, marginal profit and marginal average costs in terms of output (q). [2 marks] Marginal Revenue, Marginal Cost, Marginal Profit and Marginal Average Costs is the derivative of Revenue, Cost, Profit, Average Costs . Since the derivative of f(x) = xn is nxn-1, we have: Marginal Revenue = 575 - q Marginal Cost = 3q2 -50q Marginal Profit = -3q2 + 49q +575 Marginal Average Cost = 2q - 25 -45,000/q2 (since 1/q = q-1) (c) Find the output levels of the firm that and confirm that the output levels found do indeed maximise or minimise these functions [ 1 mark] (i) Maximise revenue To maximise revenue, set marginal revenue to 0 q = 575 This is the graph of Revenue = 575q - q2 , we can see that it is maximised at q = 575. (ii) Minimise costs To minimise costs, set marginal costs to 0 q = 50 / 3 or approx 17 units This is the graph of Costs = q3 - 25q2 + 45,000. We can see that the minimise value is approximately at q =17. (iii) Maximise profits To maximise profits, set marginal profits to 0 -3q2 + 49q +575 = 0 Using the quadratic formula, we have: q = 23.23 , -7.89 Disregarding the negative value, we have: q = 23 units. This is the graph of Profit = -q3 + 24.5q2 + 575q - 45,000. We can see that the maximum value is approximately at q=23. (iv) Minimise average costs To minimise average costs, set marginal average costs to 0: 2q - 25 -45,000/q2 = 0 (multiply both sides by q2) 2q3 - 25q2 - 45,000 = 0 With the use of trial and error, we get the only possible value as: q = 33 units. This is the graph of Average Cost = q2 - 25q + 45,000/q. We can see that the maximum value is approximately at q=33. 2. The demand function for a product is given by the following expression: q = 25 + 200 (p - 2) (a) Calculate the demand at prices 3 and 7 [1/2 mark ] For p = 3: q = 25 + 200 (3 - 2) q = 25 + 200 q = 225 For p = 7: q = 25 + 200 (7 - 2) q = 25 + 40 q = 65 Answer in (Q,P) form: (225,3), (65,7) (b) Calculate the ARC elasticity of demand with respect to price between the prices given in part (a) and comment on whether demand is elastic or inelastic between these prices. [1/2 mark] Earc = (Q2-Q1) / [(Q2+Q1)/2] (P2-P1) / [(P2+P1)/2] Earc = (65-225) / [(65+225)/2] (7-3) / [(7+3)/2] Earc = -160 / 145 4 / 5 Earc = -40 = -1.38 29 Since an "elastic" good is where price elasticity of demand is greater than one, we can consider that the demand is elastic between these prices. (c) Find an expression for POINT elasticity of demand with respect to price in terms of price. [ 1 mark] Ept = (q/ p) * p/q The derivative of q = 25 +200/(p-2) is q/ p = 0 + -1 (200) (p-2)-2 And q = 25 +200/(p-2) Hence: Ept = [-200p/ (p-2)2]/ [25 +200/(p-2)] (d) Calculate POINT elasticity of demand at prices 3 and 7 and comment on their values and on the relationship between ARC and POINT elasticity [1/2 mark] Ept = [-200p/ (p-2)2]/ [25 +200/(p-2)] Ept (3) = (-600/ 1)/ 225 = -2.67 Ept (7) = -56/ 65 = -0.862 The value of arc elasticity is in between the value of point elasticity which is expected. (d) What price(s) give unit POINT elasticity [1/2 mark] Ept = [-200p/ (p-2)2]/ [25 +200/(p-2)] = 1 1 = [-200p/ (p-2)2]/ [25 +200/(p-2)] 1 = [ -200p (p-2) / (25p -50 + 200) (p-2)2] 1 = [ -200p / (25p +150) (p-2)] 1 = [ -200p / (25p +150) (p-2)] 25p2 +150p -50p -300 = -200p 25p2 +300p -300 = 0 p2 +12p -12 = 0 Using quadratic formula, we get: p = 0.93 and p = -12.93 Disregarding the negative price, we have p =0.93 3. The marginal cost of production for a particular product has been modelled by: MC = 30q2 - 45q - 7500 where q is the quantity produced. (a) If fixed costs of production are 25,500 find an expression for the total cost function C. [2 marks] The reverse of a derivative is an integral. Since marginal cost is an integral of cost, the general formula for the product cost is the integral qn = qn+1/ (n+1) + C Disregarding the C (constant) since no data is given to derive it: C = 30q3 / 3 - 45q2/2 -7500q C = 10q3 - 22.5q2 -7500q Total cost is then: C = 10q3 - 22.5q2 -7500q +25,500 (b) If marginal revenue for this product has been modelled by: MR = 30q2 - 75q find an expression for the revenue function R. [1/2 mark] The reverse of a derivative is an integral. Since marginal cost is an integral of cost, the general formula for the product cost is the integral qn = qn+1/ (n+1) + C Disregarding the C (constant) since no data is given to derive it: R = 30q3 / 3 - 75q2/2 R = 10q3 - 37.5q2 (c) Find an expression for profit and hence, or otherwise determine the production quantity that maximises profit. [1/2 mark] Profit = Revenue - Cost = 10q3 - 37.5q2 - (10q3- 22.5q2 -7500q +25,500) = -15q2 + 7500q - 25,500 To maximize q, we get the derivative of the profit function and set to 0: -30q + 7500 = 0 q = 250 Read More