(2x + 1) x 3 3x2 =

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1 BEEM13 Optimization Techniques for Economists Homework Week 2 Solutions Dieter Balkenborg Departments of Economics Universit of Eeter Eercise 1 Find the derivative of (2 + 1) 1 ln 1 3 Solution 1 d d (2 + 1)1 = 1 (2 + 1) 9 2 = 2 (2 + 1) 9 d d ln 1 3 = = Eercise 2 Sketch the graph of a function () that has all the following properties: i) () > when < and when > 4 ii) () < when < < 4 iii) () > when > 3 iv) () < when < 3: Solution 2 The graph of a function with the required properties is given in the rst picture below. The net two give the graphs of the rst and the second derivatives. The function has a peak at = ; a trough at = 4 and an infection point at = 3. It is hence increasing / upward sloping for < and > 4 and decreasing / downward sloping for < < 4. Correspondingl, the rst derivative is positive for < and > 4 and negative for < < 4, as can also be seen in the second graph. The function is concave for < 3 and conve for > 3. Correspondingl the second derivative is negative for < 3 and positive for > 3; as can be seen in the second graph

2 You were onl required to sketch a gtraph as above free hand. I could not resist tring to nd eplicit function with the required properties, but it took me ages and I used maple quite a lot. The problem is that a cubic function would not work because a cubic has its in ection point eactl in the middle between the peak and the trough (if the two latter eist). So, having the critical points nailed at = and = 4, the in ection point had to be at = 2. I then tried to use a polnomial of order. I assumed that its derivative would be of the form ( 4) 2 + b + c The derivative of the latter would have to have the factor ( division ields d d ( 4) 2 + b + c = (3b 12) 2 + (2c 8b) 4c 3) ; in order to have the in ection point in the right place. Polnomial (3b 12) 2 + (2c 8b) 4c 3 = b + 2c + b + 2c + 3b 3 and so we must have 2c + 3b = : In order to avoid additional in ection points I also wanted the quadratic factor b + 2c + b of the third derivative to have no real root, which works for b = 1 and c = 3 2. (Notice form the third graph above that I onl just avoid the additional roots.) I multiplied ( 4) b two to eliminate a fraction, epanded and integrated to get m function on which the above graphs are based. The constant 1 was onl added to make the graph look clearer, not going through on of the numbers at the coordinate ais etc. Eercise 3 Consider the function () = e 2 i) Calculate and draw a sign diagram for the rst derivative. Where is the function increasing or decreasing. Are there an peaks or troughs? Does the function have an (absolute) maimum. ii) Calculate and draw a sign diagram for the second derivative. Where is the function conve or concave. Are there an in ection points? 2

3 Solution a) = 2e 2. Thus the function has a unique critical point at zero. From the sign of the derivative we see that the function is increasing to the left and decreasing to the right. Hence = is the (global) maimum of the function. Let g (u) = e u and h () = 2. g (u) is increasing and h (u) is concave. As a monotone transformation of a concave function the function () = e 2 = g (h ()) is quasi concave. b) = e 2. The second derivative is zero at = p 1 2 ; in between it is negative, outside positive. = 1 p 2 are hence in ection points with the function concave in between the roots and conve outside. Eercise 4 Find all critical points of the function Are these local maima or minima? () = Solution 4 d d = 3 4 = 2 4 = ( + 2) ( 2) The function has critical points at = ; +2; 2. d 2 d 2 = 32 4 = 3 + r! 4 3 r! 4 3 The function has in ection points at = p 4=3 and at = p 4=3 < 2. It is conve for < p 4=3 and > p 4=3. It is concave in between. Hence = +2 and = 2 are local minima while = is a local 3

4 maimum. All this is visible from the graph of the function and its two derivatives: Eercise For the function = nd the maimum a) on the interval [ 2; 2] and b) on the interval [ 4; 4] : Solution We have (4) = 32 (2) = 4 () = On the interval [ 2; 2] the minima are hence at 2 and the maimum is at =. On the interval [ 4; 4] the minima are hence at 2 and the maima are at =. Eercise 6 A producer operating in a perfectl competitive market has the total cost function where costs are given in Pounds Sterling. T C (Q) = 2Q 3 18Q 2 + 6Q + 1. Calculate and sketch the marginal cost function M C (Q) and the average variable cost function AV C (Q). 4

5 2. Solve the equation MC (Q) = AV C (Q). 3. At what quantit are average variable costs minimized? 4. What are the minimum average variable costs?. What quantit maimizes pro ts when the market price is P = 1? 6. What is the maimal pro t the rm can make at this price? 7. What quantit maimizes pro ts when the market price is P = 2? Solution 6 1. MC (Q) = 6Q 2 36Q + 6 AV C (Q) = 2Q 2 18Q AVC are drawn fat. 2) MC (Q) = 6Q 2 36Q + 6 = 2Q 2 18Q + 6 = AV C (Q), 4Q 2 18Q = 4Q (Q 4:), Q = or Q = 4: 3) The minimum of AV C is at 4., where AV C and MC meet (in the positive), i.e. at Q = 4:. This can also be proved directl: dav C = 4Q 18 =, Q = 4: dq d 2 AV C dq 2 = 4 > so the function is conve Minimal average costs are AV C (4:) = 19: 4) At the price P = 1 minimal average costs cannot be covered at an quantit. Therefore zero output is optimal. The pro t function is in the case P Q T C (Q) = 1Q 2Q 3 18Q 2 + 6Q + and has the following graph

6 ) At best the rm loses its ed cost, i.e., the optimal pro t is. 6) Now the minimal average costs are covered b the price and it is optimal to produce where price equals marginal costs. This gives the equation MC (Q) = 6Q 2 36Q + 6 = 2 The solutions to this quadratic equation are Q = p 114 4:779 and Q = 3 1 6p 114 1:22. We have to take the larger root. The pro t function is in the case P Q T C (Q) = 2Q 2Q 3 18Q 2 + 6Q + and has the following graph