DSC1520 2023 S1 Assignment 5 (Mock Exam) Solutions NS Nkomo1 and Prof. S Mukeru2 1,2 University of South Africa Department of Decision Sciences Email ids: 1 nkomons@unisa.ac.za, 2 mukers@unisa.ac.za 1,2 Question 1 The surfing club provides swimming lessons to boost their funds. The club has fixed costs of R1 035 per day, when offering these classes-mostly for insuranceand it costs them R195 for each lesson given. How many lessons did they provide if total costs amounted to R3 000 ? Solution 1 The surfing club’s total costs function is T C = F C + V C = 1 035 + 195q, where q is the number of lessons given per day. When the club’s total costs is R3 000 then the number of lesson will be 3 000 = 195q 3 000 − 1 035 = 1 965 1 965 = 195 = 10, 0769 ≈ 10. q q 1 035 + 195q = Question 2 A firm’s fixed cost to produce cups amount to R1 260 per week and it costs R22 to produce each cup. These cups are sold for R65 each. What would be the firm’s profit if they produce and sell 160 cups? 1 Solution 2 prof it(P ) = T R − T C where TR = price per cup × number of cups sold = 65q, where q is the number of cups produced (or sold) per week. TC = FC + V C = 1 260 + 22q. Hence, the profit function becomes P = 65q − (1 260 + 22q) = 65q − 1 260 − 22q = 43q − 1 260. Also, when they sell 160 cups, their profit is R5 620, which can be calculated as P (160) = 43(160) − 1 260 = 5 620. Question 3 The demand and supply functions for pans are q = 500 − 5p and p = 0, 8q + 30, respectively, with the price in rand and q the quantity in boxes. Find the point(s) where the demand and supply function intersect. Solution 3 The demand function q = 500 − 5p can be transformed to p= 500 q − = 100 = 0, 2q. 5 5 Now, that both the demand and supply function are in the form p(q), we can equate them to find the point where they intesect. 100 = 0, 2q = 0, 8q + 30 100 − 30 = 0, 2q + 0, 8q q = 70. To find the value of p, we substitute q = 70 on one of the given equations. Then p = 100 − 0, 2(70) = 86. 2 Question 4 The demand function for a certain kind of cellphones is given by q = 90 000 − 5p. Determine the point elasticity of demand when p = 5 000. Solution 4 When p = 5 000, then q = 90 000 − 5(5 000) = 65 000, that is (p0 ; q0 ) = (5 000; 65 000) and p0 q0 5 000 = −5 × 65 000 = −0, 385. = −b × ϵd |ϵ| = 0, 385 < 1, the demand is inelastic at this price. 1% increase (or decrease) in price will cause a 0, 385% decrease (or increase) in demand. Question 5 Consider the supply function for a kind of computer, namely p = 10 + 0, 5q. Determine the arc elasticity if price increases from R1 000 to R3 000. Solution 5 The supply function p = 10 + 0, 5q can be transformed to q = 2p − 20. We need to find the quantities demanded at these prices. If p = 1 000, then q = 2(1 000) − 20 = 1 980, to get (p1 ; q1 ) = (1 000; 1 980). If p = 3 000, then q = 2(3 000) − 20 = 5 980, to get (p1 ; q1 ) = (3 000; 5 980). The arc price elasticuty is ϵs = = ≈ p1 + p2 q1 + q2 1 000 + 3 000 2× 1 980 + 5 980 1. q× |ϵ| = 1, the supply is unit elastic. The 1% change in supply is equal to 1% change in price. 3 Question 6 The demand and supply functions for a certain product are given as qd = 300 − 2pd and qs = −200 + 3ps , respectively. Determine the price (p) and quantity (q) at equilibrium. Solution 6 To find the equilibrium price and quantity, we need to solve the demand and supply functions simultaneously. Since the functions are both given in the form q(p), we set them equal and solve for p (the equilibrium price), that is qd = qs 300 − 2p = −200 + 3p 300 + 200 = 2p + 3p 500 = 5p p = 100. To find the equlilbrium quantity, we substitute p = 100 into either one of the given functions, to find q = 300 − 2(100) = 100 or q = −200 + 3(100). Then we have (p; q) = (100; 100). Question 7 Find the slope of the curve y = 3x3 + 4x2 − 7x + 9 at x = 10. Solution 7 The slope of the curve y = 3x3 + 4x2 − 7x + 9 is called its derivative and is given by dy = f ′ (x) = 9x2 + 8x − 7. dx The slpe at the point x = 10 is f ′ (x) = 9(10)2 + 8(10) + 7 = 973. 4 Question 8 Find H ′′ (x) where H(x) = xe2x Solution 8 H ′ (x) dxe2x dx de2x dx = x + e2x dx dx = 2xe2x + e2x · 1 = = (2x + 1)e2x Using a product rule a second time, we get H ′′ (x) = d(2x + 1)ex dx de2x d(2x + 1) (2x + 1) + e2x dx dx 2(2x + 1)e2x + 2e2x = 2(2x + 2)e2x = = Question 10 Find the stationary points of Q = p3 − p2 − 5p. Solution 10 The stationary points are when the derivative is zero, so we need to find Q′ and set it equal to zero. dQ = Q′ = 3p2 − 2p − 5 dp Now, let it equal to zero 3p2 − 2p − 5 = 0. We need to solve this quadratic equation: (3p − 5)(p + 1) = 0 Solving this gives p = 35 , p = −1. By substituting our p values 35 and −1 into Q = p3 − p2 − 5p, we get 53 , − 175 27 and −1, 3, which they are stationary points. 5 Question 11 Given the function g(x) = 1 1000 x − x3 + 1, 5 what is the third derivative of g? Solution 11 To get to the third derivative of g, we first need to find the first and then the second derivative of g. Where the fist derivative of g is given by: g ′ (x) = = dg(x) dx 200x999 − 3x2 and the second derivative of g is given by g ′′ (x) dg ′ (x) dx 199 800x998 − 6x = = Then, the third derivative of g is: d3 g dx3 = = dg ′′ (x) dx 199 400 400x997 − 6. Question 12 Find the turning point(s) of the following function and determine their nature: M (n) = n2 − 4n + 4. Solution 12 We start by differentiating the given function and then set the result equal to zero, that is M ′ (n) = 2n − 4 = 0. When we solve this equation, we find n = 2 with M (2) = (2)2 − 4(2) + 4 = 0. The turning point is therefore at (2; 0). We now need to determine whether the point is a maximum or minimum point, by using the second derivative, namely M ′′ (n) = d(2n − 4) = 2. dn Since M ′′ (2) = 2 > 0, we have a local minimum point at n = 2. 6 Question 13 Determine the intervals along which the following function increases or decreases: F (x) = −x3 + 9x2 − 24x + 26. Solution 13 To find the stationary points of F , we find the derivative of F , set it equal to zero and solve for x. The derivative is dF dx = −3x2 + 18x − 24. F ′ (x) = Now, set it equal to zero and solve −3x2 + 18x − 24 = 0 x − 6x + 8 = 0 (x − 4)(x − 2) = 0 2 x = 2 or x = 4. Therefore, F has stationary points at x = 2 and x = 4. We now need to determine whether these points are maximum or minimum points, by using the second derivative, namely F ′′ (x) = d(−3x2 + 18x − 24) = −6x + 18. dx Since F ( 4) = −6(4) + 18 = −6 < 0, we have a local maximum point at x = 4. Since F ( 2) = −6(2) + 18 = 6 > 0, we have a local minimum point at x = 2. With this information at hand, we can easily draw a rough graph of F and specify that the function F (p) decreases over (−∞; 2) and (4; ∞), while increases over (2; 4) Question 14 Suppose we have the demand function given as p = 72 − 7, 5q, where q is the number of units to be produced and sold. Determine the marginal revenue after 4 units have been sold. 7 Solution 14 From the given demand function p = 72 − 7, 5q, we find the total revenue function to be T R(q) = pq = (72 − 7, 5q)q = 72q − 7, 5q 2 . By differentiation, we find the marginal revenue function to be MR dT R dq d = (72q − 7, 5q 2 ) dq = 72 − 15q. = From the M R function, we find that when 4 units have been sold, the revenue from the next unit sold will be M R(3) = 72 − 15(4) = R12 Question 15 Given total cost function (T C) as T C = q 3 − q 2 + 120q, where q is the number of units to be produced. Determine the cost to produce the 17t h unit. Solution 15 Given Total cost function T C = q 3 − q 2 + 120q, MC dT C dq d 3 = (q − q 2 + 120q) dq = 3q 2 − 2q + 120. = When q = 16, then M C(16) = 3(16)2 − 2(16) + 120 = 856. The cost to produce the 17th unit, is R856. 8 Question 16 Determine the intervals along which the given function m(n) incraeses or decreases, where m(n) = 30n2 − n3 . Solution 16 To find the stationary points of m, we find the derivative of m, set it equal to zero and solve for n. The derivative is m′ (n) = 60n − 3n2 . Now, set it equal to zero and solve 60n − 3n2 = 0 2 = 0 n(20 − n) = 0 20n − n n = 0 or n = 20. Therefore, m has stationary points at n = 0 and n = 20. We now need to determine whether these points are maximum or minimum points, by using the second derivative, namely m′′ (n) = = d (60n − 3n2 ) dn 60 − 6n. Since m′′ (0) = 60 > 0, we have a local minimum point at n = 0. Since m′′ (20) = −60 < 0, we have a local maximum point at n = 20. With this information at hand, we can easily draw a rough graph of m and specify that m is decreasing over (−∞; 0) and (20; ∞), while increases over (0; 20). Question 17 The demand function of a firm is Q = 900 − 1, 5P, where P and Q represent the price and quantity, respctively. At what price is revenue T R at maximum? Solution 17 TR = Q×P = 900 − 1, 5P P = 900P − 1, 5P 2 . 9 To find the price at maximum T R, we need to first find T R′ . let T R′ = 0 and solve to P T R′ = −3P + 900 −3P + 900 = 0 3P = 900 P = 300 Therefore, P = 300 when the revenue is at maximum. Question 18 If the production function is given by √ Q = 400 L − 5L, where Q denotes output and L denotes the size of workforce, calculate the value of marginal production of labour if L = 16. Solution 18 Marginal product M P MP 400 dQ = √ −5 dL 2 L 200 √ − 5. L = = If L = 16, then MP = = 200 √ −5 16 45. Question 19 A firm has the following total and cost functions, respectively. TR = 10Q − 4Q2 TC = 16 − Q2 , where Q is the number of unites produced and sold (in thousands). How many unites should be produced to maximise the profit? 10 Solution 19 When the profit is at maximum: Marginal revenue (M T ) = Margigal cost (M C) MR dT R dQ 10 − 8Q = = and MC dT C dQ −2Q = = now we have MR = MC 10 − 8Q = −2Q Q = 1, 67 Question 9 The demand for seats at a mini soccer match is given by Q = 150 − P 2 , where Q is the number of seats and P is the price per seat. Find the price alasticity ϵ of demand if seats cost R5 each. What does this value mean? Solution 9 Q(5) ϵd = 150 − (5)2 = 150 − 25 = 125 = △Q P △P Q = −2 × = −0.08 |ϵd | = 0.08 < 1 11 5 125 elastic Question 21 The demand function Q and total cost function T (q) of a commodity are given by the equations P = 100 − 4q, TC = 144 + 52q, where P and q are the price and quantity, respectively. Determine the quantity that must be produced and sold at breaks even. Solution 21 From the given functions, we find the total revenue function to be T R = pq = (100 − 4q)q = 100q − 4q 2 . We know that break-even occurs when total revenue is equal to total cost. We can therefore find the break-even points for the firm by setting T R = T C, that is 100q − 4q 2 = 144 + 52q 2 −4q + 48q − 144 = 0 q 2 − 12q + 36 = 0 (q − 6)(q − 6) = 0 q = 6. The firm therefore breaks even when 6 units of the good are produced and sold. Question 22 The total cost of producing a certain good is given by T C = 30 ln (q + 300) + 150. Find the marginal cost (M C) and the avarage cost (AC) functions. Solution 22 MC = = = = dT C dq d 30 ln (q + 300) + 150 dq 30 (1) q + 300 30 q + 300 12 and AC TC q 30 ln (q + 300) + 150 q = = Question 23 Evaluate Z 1 9 √ 3t2 + t2 t − 1 dt. t2 Solution 23 We start by simplifying, √ Z 9 2 Z 9 1 3t + t2 t − 1 dt = 2 + t 2 − t−2 dt 2 t 1 1 #9 " 3 t−1 t2 = 3t + 3 − −1 2 1 #9 " −1 t 2 3 = 3t + t 2 − 3 −1 1 3 3 2 (9)−1 2 = 3 · 9 + (9) 2 − − 3 · 1 + (1) 2 + (1)−1 3 −1 3 1 2 = 27 + 18 + − 3 − − 1 9 3 = 40 Question 24 Find Z √ 2x + 5e5x dx 1 − 4x2 Solution 24 Z √ 2x + 5e5x dx 1 − 4x2 Z = √ x dx + 1 − 4x2 let u = 1 − 4x2 . Then du = −8xdx, so xdx = − 81 du, and let 13 Z e5x dx v = 5x. Then dv = 5dx, so dx = 15 dv. Therefore, Z 2x 5x √ + 5e dx = 1 − 4x2 Z Z x 2 √ dx + 5 e5x dx 1 − 4x2 Z Z 1 1 √ du + ev dv = − 4 u Z Z 1 1 u− 2 du + ev dv = − 4 1 √ = − (2 u) + ev + c 4 1 p = − ( 1 − 4x2 ) + e5x + c 2 Question 25 Find the area under the curve F (x) = x2 x , −1 covered between x = 2 and x = 4. Solution 25 The area between the curve F (x) and the x-axis from x = 4 and x = 8 is determine as Z 4 x dx. 2−1 x 2 Let u = x2 − 1, then du = 2xdx, so xdx = 12 du. If x = 4 then u = 15. If x = 8 then u = 63. Therefore, Z 4 Z x 1 63 1 dx = du 2 2 15 u 2 x −1 i63 1h ln u = 2 15 1 = ln 63 − ln 15 . 2 Question 26 The marginal cost for a certain product is given by 100 MC = , q + 30 where q is the number of units produced. Given that the total cost of producing 10 units is R500. Determine the total cost function. 14 Solution 26 Total cost (T C) can be calculated by integrating the marginal cost function, that is Z TC = M Cdq Z 100 = dq q + 30 Z 1 = 100 dq q + 30 = 100 ln (q + 30) + c. Given that T C = 500 when q = 10, we therefore find the constant, c, by setting 100 ln (40) + c = 500 to get c = 500 − 100 ln (40) = 131. The total cost function is therefore T C = 100 ln (q + 30) + 131. Question 27 At a small tool-hire company, the estimated rate of increase in the maintenance cost of power drills is given by C(t) = 2e2t + 2t + 19, where C is in rand and t is time in weeks. Give the total maintenance cost during the first 15 weeks. Solution 27 Total maintenance costs during the first 10 weeks (0 ≤ t ≤ 15) are given by Z TC = 15 2e2t + 2t + 19 dt 0 = h i15 e2t + t2 + 19t = e30 + 225 + 285 − 1 = 509 + e30 0 15 Question 28 The demand and supply functions for bikes are pd = 900e−0.1q and ps = 3e0.9q , respectively. Where p is the price, and q the quantity. What is the total surplus at equilibrium market? Solution 28 We find the equilibrium price and quantity by setting pd = ps , that is 900e−0.1q = 3e0.9q −0.1q = e0.9q 300e 300 q = eq = ln 300 To get p, we can substitute q = ln 300 into any of the given functions. p = 3e0,9 ln 300 = 508, 7804134 ≈ 509. Now, we have (p0 ; q0 ) = (509; ln 300), we can then use the consumer and concumer surplus formula, then Z ln 300 CS = 900e−0.1q dq − (509)(ln 300) 0 900 −0,1q ln 300 e − 2901 0 (−0, 1) =−0,1 ln 300 = −9 000 e − 1 − 2901 = = 1 009 and Z PS = ln 300 (509)(ln 300) − 3e0.9q dq 0 = 2 901 − = 2 341 i 3 h 0,9 ln 300 e −1 0, 9 We can not find the total surplus (T S), that is TS = CS + P S = 1 009 + 2 341 = 3 350 16 Question 29 The marginal cost of a good is given by M C = −10q 4 + 3q 2 + 80q, where q is the number of units produced. It is also given that the fixed cost of production is R1 500. Determine the total cost function. Solution 29 We can find the total cost by integrating the marginal cost, that is Z TC = M Cdq Z 4 2 = − 10q + 3q + 80q dq = −2q 5 + q 3 + 40q 2 + c, given that F C = 1 500, this means T C(0) = 1 500, that is 1 500 = −2(0)5 + (0)3 + 40(0)2 + c therefore, c = 1 500, then T C = −2q 5 + q 3 + 40q 2 + 1 500 Question 30 Suppose the marginal revenue function for a good is M R = 100 − 13q 12 + 16q 3 , where q is the number of units sold, and that total revenue from selling one unit is R3 316. Find the total revenue function. Solution 30 We can find the total revenue by integrating the marginal reveue, that is Z TR = M Rdq Z = 100 − 13q 12 + 16q 3 = 300q − q 13 + 4q 2 + c. Given that total revenue from selling one unit is R3 316, meaning T R(1) = 3 316, that is 3 316 = 100(1) − (1)13 + 4(1)2 + c therefore, c = 3 213, then T R = 100q − q + 4q 2 + 3 213. 17