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Unformatted text preview: CHAPTER 8 Unconstrained Optimisation Problems with One or More Variables In Chapter 6 we used differentiation to solve optimisation problems. In this chapter we look at more economic applications of this technique: profit maximisation in perfect competition and monopoly ; and strategic optimisation problems that arise in oligopoly or when there are externalities . Then we show how to maximise or minimise a function of more than one variable , and again look at applications. ./ 1. The Terminology of Optimisation Suppose that an economic agent wants to choose some value y to maximise a function ( y ). For example, y might be the quantity of output and ( y ) the profit of a firm. From Chapters 5 and 6 we know how to solve problems like this, by differentiating. The agents optimisation problem is: max y ( y ) ( y ) is called the agents objective function- the thing he wants to optimise. To solve the optimisation problem above, we can look for a value of y that satisfies two conditions: The first-order condition: d dy = 0 and the second-order condition: d 2 dy 2 < But remember that if we find a value of y satisfying these conditions it is not necessarily the optimal choice, because the point may not be the global maximum of the function. Also, some functions may not have a global maximum, and some may have a maximum point for which the second derivative is zero, rather than negative. So it is always important to think about the shape of the objective function. An optimisation problem may involve minimising rather than maximising for example choosing output to minimise average cost: min q AC ( q ) The first- and second-order conditions are, of course, dAC dq = 0 and d 2 AC dq 2 > 0, respectively. 135 136 8. UNCONSTRAINED OPTIMISATION PROBLEMS 2. Profit Maximisation Suppose a firm has revenue function R ( y ) and cost function C ( y ), where y is the the quantity of output that it produces. Its profit function is: ( y ) = R ( y )- C ( y ) The firm wants to choose its output to maximise profit. Its optimisation problem is: max y ( y ) The first-order condition for profit maximisation is: d dy = R ( y )- C ( y ) = 0 which is the familiar condition that marginal revenue equals marginal cost . Examples 2.1 : A monopolist has inverse demand function P ( y ) = 35- 3 y and cost function C ( y ) = 50 y + y 3- 12 y 2 . (i) What is the profit function? ( y ) = yP ( y )- C ( y ) = (35- 3 y ) y- (50 y + y 3- 12 y 2 ) =- y 3 + 9 y 2- 15 y (ii) What is the optimal level of output? The first-order condition is: d dy =- 3 y 2 + 18 y- 15 = 0 There are two solutions: y = 1 or y = 5. Check the second-order condition: d 2 dy 2 =- 6 y + 18 When y = 1, d 2 dy 2 = 12 so this is a minimum point....
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- Spring '11