Section A (Basic algebra and calculus multiple choice)

Transcription

1 BEE1 Basic Mathematical Economics Dieter Balkenborg January 4 eam Solutions Department of Economics University of Eeter Section A (Basic algebra and calculus multiple choice) Question A1 : The function f () = ( ( 2) 2 (+2) (+2)( 2) for > and6= 2 4 for =2 is h1i not well-defined. h2i conve. h3i concave. h4i discontinuous. h5i not differentiable. h6i decreasing. Solution A1 : h2i Question A2 : Which of the following functions is non-negative on the interval 2 < <2: h1i y = 3 ( 3 1) h2i y = h3i y = 2 ( 2 4) h4i y = ( 4) ( +4) h5i y = 4 ( 4 1) h6i y = conve. Solution A2 : h2i

2 Question A3 : The inequality ( 2) 2 ( 1) 2 ( +1)( +2) canbesolvedas h1i 2or 1 or1 2 h2i 2 1and =and1 2 h3i 2 <<1and =and1<<2 h4i 2 <<1or <2 h5i 2and 1 and 1and =2 h6i 2or 1 1or =2 Solution A3 : h6i Question A4 : A consumer has spent $1, on designer clothes. Current VAT is 17.5%. How much VAT did the consumer pay. h1i $175 h2i $17.5 h3i $ h4i $ h5i$. h6i Solution cannot be determined because of insufficient information. Solution A4 : h4i Question A5 : Which statement is correct for the function f () = n with n an integer? h1i If n is odd and positive then =isapeak. h2i If n is odd then the function is concave on the whole real line. h3i If n is odd and negative then the function is concave on the negative numbers. h4i If n is even then the function is conve on the whole real line. h5i If n is even and negative then the function is both positive- and negative-valued. h6i If n is even and positive then the function is non-decreasing. Solution A5 : h3i 2

3 Question A6 : You are given the following information about the signs of the first and second derivative of a function y (): y () + + y () Which of the following si graphs is consistent with this information? Graph 1 Graph 2 Graph Graph 4 Graph 5 Graph 6 Solution A6 : h3i Question A7 : The following statement is false for the function f () = 4 : h1i On the interval 2 3 the function has a minimum at =. h2i Thefunctionhasauniquemaimumontheinterval 2 3. h3i The function is strictly conve. h4i The function has an inflection point at =. h5i The function is decreasing on the non-negative numbers. h6i Thefunctionhasauniqueroot. Solution A7 : h4i 3

4 Section B (Basic algebra and calculus eercise) Question B1 : Simplify ( 2) ( 9 +3)( +2) ( 2) ( )( +2) Solution B1 : 2 ( 2) ( +2)=8 2 3 Question B2 : Simplify (52 y 3 ) 2 (1 3 y 2 ) 3 Solution B2 : ( 5 2 y 3 ) 2 (1 3 y 2 ) 3 (y 2 ) 3 (y 2 ) 3 = 254 y 6 9 y 6 3 y 6 =25 6 =25 (1 4 5 ) Question B3 : Consider the function y = on the domain { }. Drawthe graph of the function, Is the function invertible. If yes, what is the inverse function? Solution B3 : The function is invertible. The inverse is given by y =, = y = 2+ 2+y. (We must take the larger root.) Question B4 : Find all solutions to the quadratic equations i) = 2 ii) = iii) =8 q q Solution B4 : i) = 1 ± 1 1 (no real solution) ii) = 1 ± q q = 1 ± 1 +2= 1 ± 9 =1or =or 1 iii) Question B5 : Solve the simultaneous systems of equations i) 8 = 2 +6y, 9y =12 6 ii) 8 = 4 +6y, 9y =12 6 Solution B5 : i) =,y =1 1 3 ii) arbitrary, y = Question B6 : Solve the simultaneous system of equations 2 + y 2 =5,y +2 =. Solution B6 : =5 2 =5, = ±1, y = 2. Question B7 : Use a sign diagram to find all solutions to the inequality ( +1) 2 ( 2 4) <. 4

5 Solution B7 : y =( +1) 2 ( 2 4) = ( +1)( +2)( 2) < 2 = 2 2 << 1 = 1 1 <<2 2 2 < ( +1) y + + Thus y<if 2 << 1 or 1 <<2. Question B8 : Factorize the polynomial y = Solution B8 : =1isaroot q = has the solutions = 1 ± = 1 or 3. This quadratic polynomial has hence the factorizations =(2 +3)(2 1). Overall y = =( 1) (2 +3)(2 1) Question B9 : Consider the first, second, third and so on derivative f (n) () ofthe polynomial 4 +. Fromwhichn onwards will all derivatives f (n) () be constant? Solution B9 : f =4 3 +1, f =12 2, f =24, f (iv) =24,f (v) = f (vi) =... =. The fourth and all higher derivatives are constant. Question B1 : What is the slope of the tangent to the graph of the function y = at =2? Solution B1 : y = y (2) = 3 ( ) 2 (12 4+1)= =675 Question B11 : Is =aninflection point of the function y = 5 + 3? Solution B11 : Yes the second derivative y = 3 +6 =( 2 +6) changes sign at. 5

6 Question B12 : Differentiate the function y = You can obtain one additional mark for simplifying. Solution B12 : y = (2 +1)(2 +1) ( )(2 1) ( 2 +1) 2 = ( 2 +1) 2 Section C (Applied calculus) Question C1 : cost function A producer operating in a perfectly competitive market has the total where costs are given in Pounds Sterling. TC(Q) =2Q 3 12Q 2 +3Q +5 C1-i : Calculate and sketch the marginal cost function MC (Q) and the average variable cost function AV C (Q). (1 mark) Solution C1-i: MC (Q) =6Q 2 24Q +3,AV C (Q) =2Q 2 12Q Q C1-ii : Solve the equation MC (Q) =AV C (Q). (1 mark) Solution C1-ii: 6Q 2 24Q +3 = 2Q 2 12Q +3 4Q 2 12Q =4Q (Q 3) = Q =orq =3 C1-iii : At what quantity are average variable costs minimized? What are the minimum average variable costs? (2 marks) 6

7 Solution C1-iii: Average costs are minimized at Q = 3, where AVC curve intersects MC. AV C (3) = =12. C1-iv : What quantity maimizes profits when the market price is P = 1? What is the maimal profit thefirm can make at this price? (2 marks) Solution C1-iv: Price is below minimal AVC, so Q =is optimal C1-v : What quantity maimizes profits when the market price is P =? (1 mark) Solution C1-v: The profit maimizing quantity is the larger solution Q = 5 tomc (Q) = 2Q 2 12Q +3=. Solutionis: Q =5. Question C2 : A manufacturer can produce 4 wardrobes at the cost of $37, and 7 wardrobes at the cost of $55,. Assuming a linear cost function, what are his marginal costs and what are his fied costs? (3 marks) Solution C2: Building 3 wardobes more costs $18. more. The marginal costs are hence $6. This gives a variable cost of $24,tobuild4wardrobes.Thefied costs are hence $13,. Question C3 : A manufacturer thinks that the economy is in a slump and he has to cut prices. He has been selling lamps at $7 apiece, and at this price consumers have been buying 6, lamps per month. The manufacturer estimates that for each $1 reduction in price, 6, additional lamps can be sold each month. The manufacturer can produce at costs $2 per lamp. At what price should the manufacturer sell the lamps to generate thegreatestpossibleprofit? (4 marks) Solution C3: Π ( p) =(7+ p 2) (6 6 p) = 6 (5 + p)(1 p) roots at p = 5 and p = 1, optimal price increase in the middle at p = 2. Thus price reduction by $2 optimal. 7

8 Section D (Economic Applications) Question D1 : D2-i : A price taking producer has the production function Q = L. The costs of the labour input L aregivenbythewageratew>. If the output price is normalized to 1, what is the maimal profit Π (w) of the producer as a function of the wage rate? (4 marks) h L Solution D1-i: Π (w) =ma L π (L, w) =ma L wl i,foc:π = 1 2 w =, L L = 1, Π (w) =π 1,w = 1 w 1 = 1. 2w 4w 2 2w 4w 2 4w D2-ii : Given Π (w)calculateforeachl>the minimal value G (L) =min w> Π (w)+ wl. What function do you obtain? Can you give an economic interpretation of the result? (4 marks) 1 Solution D1-ii: G (L) =min w> H (w, L) =min w> Π (w) ³ +wl,l =min w> + wl, 4w FOC: 1 + L =,w 2 = 1, w = 1 4w 2 4L 2, G (L) =min 1 L w> H 2 L L 2 = L L = L Question D2 : A monopolist faces the demand D (p) =1 p units of a commodity at price p which costs him $2 per unit to produce. Determine without using calculus the optimal profit and verify using only algebra that this is indeed the optimal profit. (4 marks) Solution D2 : The profit Π (p) =(p 2) (1 p)iszeroatp =2andp = 1. Therefore it is maimized in the middle at p =6. Indeed,Π (p) = p 2 +12p = (p 6) = 16 (p 6) 2 is maimized at p =6withprofit 16. (BEE1 January 4) (End of solutions.) 8