Topic 8 Unconstrained Optimisation: Functions of Several Variables 1 . Find the stationary points of the following functions, and determine whether each is a maximum, minimum or saddle point. (a) z x 2 y 2 4y 3x 50 First Order Necessary Condition for a stationary point: set first partial derivatives equal to 0, so Zx = 0 and Zy = 0 Then, taking first derivatives and setting them equal to zero, we have zx 2x 3 0 (1) zy 2y 4 0 (2) Solving these equations we get: From (1), x 32 ; from (2), y 2 . So there is a stationary point at (- 3/2 , 2) Second Order Sufficient Condition for a maximum/minimum/saddle point: MAX: Zxx < 0 AND Zxx Zyy – (Zxy)2 >0 MIN: Zxx > 0 AND Zxx Zyy – (Zxy)2 >0 SADDLE: Zxx Zyy – (Zxy)2 < 0 Checking the second derivatives: Zxx = – 2 < 0 at all values of x (and hence when x = - 3/2) Zyy = – 2 Zxy = Zyx = 0 Checking Zxx Zyy – (Zxy)2 >0 when x = - 3/2 and y = 2 we have ( - 2 . - 2) – (0)2 > 0 Confirming that (- 3/2 , 2) is a maximum (b) z 2x 5y 5x 2 2y 24xy 1000 1 First Order Necessary Condition for a stationary point: set first partial derivatives equal to 0, so Zx = 0 and Zy = 0 Then, taking first derivatives and setting them equal to zero, we have zx 2 10 x 4y 0 (1) zy 5 4y 4x 0 (2) Solving these equations we get: 4y = + 2 – 10x = + 5 – 4x Thus, 6x = - 3 and so x = - ½ Hence from eq (1), y = ½ - 5/2 x = ½ - (5/2 . – ½ ) = ½ + 5/4 = 7/4 Second Order Sufficient Condition for a maximum/minimum/saddle point: MAX: Zxx < 0 AND Zxx Zyy – (Zxy)2 >0 MIN: Zxx > 0 AND Zxx Zyy – (Zxy)2 >0 SADDLE: Zxx Zyy – (Zxy)2 < 0 Checking the second derivatives: Zxx = 10 > 0 at all values of x (and hence when x = - ½ ) Zyy = 4 Zxy = Zyx = 4 Checking Zxx Zyy – (Zxy)2 >0 when x = - ½ and y = 7/4 we have ( 10 . 4 ) – (4)2 > 0 Confirming that this is a minimum 2 2. A firm charges different prices for its good to its domestic and export markets. In the domestic market its demand function is given by PD 20 2QD while in the export market its demand function is given by PE 15 1.5QE . The total cost function is TC 5QD QE . i) What level of output should the firm sell in each market in order to maximise profit? [hint: total profit = Total Revenue Domestic market + Total Revenue Export Market - Total Cost ] ii) What price will the firm charge in each market if it is to maximise its profits? The first step is to find an expression for the profit function for the firm. To do this we need to find an expression for total revenue. Total revenue in the domestic market will be given by: TRD PD QD 20 2QD QD 20QD 2QD2 Total revenue in the export market will be given by: TRE PE QE 15 1.5QE QE 2 15QE 1.5QE2 Total revenue from sales in both markets will be given by: TR TRD TRE 20QD 2QD2 15QE 1.5QE2 Total cost is given by TC 5QD 5QE Profit is therefore given by TR TC 20QD 2QD2 15QE 1.5QE2 5QD 5QE 15QD 2QD2 10QE 1.5QE2 First order conditions: 15 4Q D 0 Q D thus, 4QD 15 and hence QD 15 4 10 3Q E 0 Q E thus, 3QE 10 and hence QE 10 3 3 Second order conditions: 2 4 0 QD2 suggests a max, but need to check second condition 2 3 0 QE2 2 0 Q D Q E 2 Q 2 D 2 . Q 2 E 2 Q Q D E 2 2 4 . 3 0 12 0 therefore we have a Maximum at QD 15 4 and QE 10 3 . These describe the levels of output the firms should sell into each market in order to maximise profits ii) In order to find the price, substitute optimal QD and QE into demand function to find values for price in each market. PD 20 2QD 20 215 4 12.5 PE 15 1.5QE 15 1.510 3 10 3. A firms inverse demand curve is described by P 40 2 A 2Q where A is expenditure on advertising, P is the price of its product and Q is the quantity demanded. Total costs of production are given by TC Q 2 AQ A 2 . What levels of output Q and advertising expenditure A should the firms choose in order to maximise profit? i) objective: maximise profit subject to Q and A TR TC TR P.Q TR 40 2 A 2Q .Q 40Q 2 AQ 2Q 2 40Q 2 AQ 2Q 2 Q 2 AQ A 2 40Q AQ 3Q 2 A 2 4 First order conditions: 40 A 6Q 0 Q Q 2A 0 A (1) (2) From equation (1) we find that Q 2 A Substitute this into equation (2) and we get: 40 A 62 A 0 40 11A 0 40 11A A 11 40 now substitute this into Q = 2A and we find: Q 2 A 211 40 22 40 11 20 One stationary point at A 11 40 , Q 11 20 Second order conditions: 2 6 0 suggests a max, but we need to check the second condition Q 2 2 2 0 A 2 2 1 QA 2 2 Q 2 . A 2 2 2 2 . 2 1 11 0 QA 6 Thus, a maximum at Q 11 20 and A 11 40 5