Math Exercises Problem Example | Topics and Well Written Essays - 1000 words

Exercises - Math Problem Example1A rigid manufactures and sells q units of a product at terms = (575 q)which has unit costs of (q2 25q) and fixed costs of 45,000.(a)Write graduate expressions for revenue, profit and average cost in terms of fruit(q) of the firm.1 markRevenue = (575 q ) q = 575q q2Profit = Revenue Total personify = 575q q2 - (q2 25q )q +45,000 = 575q q2 q3 + 25q2 - 45,000 = q3 + 24.5q2 + 575q - 45,000Average 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 peripheral Revenue, fringy Cost, Marginal Profit and Marginal Average be is the derivative of Revenue, Cost, Profit, Average Costs . Since the derivative of f(x) = xn is nxn-1, we set aboutMarginal Revenue = 575 qMarginal Cost = 3q2 -50qMarginal Profit = -3q2 + 49q +575Marginal 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 play down these functions 1 mark(i)Maximise revenueThis is the graph of Revenue = 575q q2 , we can see that it is maximised at q = 575.(ii) minimize costsTo minimise costs, arrange marginal costs to 0q = 50 / 3 or approx 17 units This is the graph of Costs = q3 - 25q2 + 45,000. We can see that the minimise pass judgment is rough at q =17. (iii) Maximise profitsTo maximise profits, set marginal profits to 0 -3q2 + 49q +575 = 0Using the quadratic formula, we haveq = 23.23 , -7.89Disregarding the negative value, we haveq = 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 costsTo minimise average costs, set marginal average costs to 02q - 25 -45,000/q2 = 0(multiply both sides by q2)2q3 - 25q2 - 45,000 = 0With the use of trial and error, we get the only possible value asq = 33 units.This is the graph of Average Cost = q 2 - 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 expressionq = 25 + 200 (p - 2) (a)Calculate the demand at prices 3 and 71/2 mark For p = 3q = 25 + 200 (3 - 2) q = 25 + 200 q = 225For p = 7q = 25 + 200 (7 - 2) q = 25 + 40 q = 65Answer in (Q,P) form (225,3), (65,7)(b)Calculate the ARC crack of demand with respect to price amidst the prices given in part (a) and comment on whetherdemand is elastic or dead between these prices.1/2 markEarc = (Q2-Q1) / (Q2+Q1)/2 (P2-P1) / (P2+P1)/2Earc = (65-225) / (65+225)/2 (7-3) / (7+3)/2Earc = -160 / 145 4 / 5Earc = -40 = -1.38 29 Since an elastic good is where price cinch of demand is greater than one, we can consider that the demand is elastic between these prices.(c)Find an expression for stratum elasticity of demand with respectto price in terms of price. 1 markEpt = (q/ p) * p/qThe derivative of q = 25 +200/(p-2) isq/ p = 0 + -1 (200) (p-2)-2And q = 25 +200/(p-2)HenceEpt = -200p/ (p-2)2/ 25 +200/(p-2)(d)Calculate POINT elasticity of demand at prices 3 and 7 andcomment on their values and on the relationship between ARC and POINT elasticity1/2 markEpt = -200p/ (p-2)2/ 25 +200/(p-2)Ept (3) = (-600/ 1)/ 225 = -2.67Ept (7) = -56/ 65 = -0.862The value of arc elasticity is in between the value of point elasticity which is expected