DSC1520Quantitative Modelling 1

Quantitative Modelling 1 DSC1520 Question 1 1. (2a 2 b 3 ) 2 × (ab4 ) 3 (2a 3b 2) 4 = (4a 4 b 6 ) × (a 3 b 12) 16a 12b 8 = a 4+3−12b 6+12−8 4 = a −5 b 10 4 = b 10 4a 5 . 2. q 4x y−4 = √ 4x × p y 4  = (4x) 1 2 × y 4 ) 1 2 = (4 1 2 x 1 2 × y 2 ) = 2y 2√ x Question 2 1. −x 2 + 8x − 16. For roots we make use of the quadratic formula x = −b − √ b 2 −4ac 2a and x = −b + √ b 2 −4ac 2a x = −8 ± q (8)2 −4(−1)(−16) 2(−1) = −8 ± √ 0 −2 = 4. 2. y = −x 2 + 8x − 16 ⇒ a = −1, b = 8, c = −16 x = −b 2a = −(8) 2(−1) = −8 −2 = 4 units Question 3 1. Revenue or Income is defined as price times quantity or R = p × q or p × x. Now given is quantity as x and price is given as p(x) = 10 − x 1000 . Thus Revenue = p × x = (10 − x 1 000 ) × x = 10x − x 2 1 000 or 10x − 0,001x 2 . 3 This study source was downloaded by from CourseH on :58:30 GMT -06:00 This study resource was shared via CourseH 2. (a) Profit = revenue − cost Price = − −b 2a = −10 2(− 1 1000 ) = 5 000 or Price = (10 − x 1 000 ) = R5 (b) Maximum revenue = (10 − 5 000 1 000 )(5 000) = 25 000. (c) The number of radios to produce to realize maximum profit profit = revenue-cost = 10x − x 2 1 000 − 2x − 5 000 = − x 2 1 000 + 8x − 5 000. number of radios = − −b 2a = −8 2(− 1 1000 ) = 4 000. (d) Maximum profit = −( 4 0002 1 000 ) + 8(4 000) − 5 000 = 11 000 (e) The price that the company should charge for each radio to realize maximum profit At maximum profit, p = 10 − 4 000 1 000 = 6. Question 4 1. Given the function f(x) = −2x 2 + 10x − 8. To graph the function we need to determine the vertex, roots and y-intercept of the function. The graph has a maximum point since a 0. The y-intercept is −8. Vertex is the point: x = −b 2a = −10 2(−2) = 2,5 y = −2(2,5)2 + 10(2,5) − 8 = 4,5. The roots are: x = −b ± √ b 2 −4ac 2a = −(10) ± q (10)2 −4(−2)(−8) 2(−2) = −10 ± 6 −4 = 1 or 4 4 This study source was downloaded by from CourseH on :58:30 GMT -06:00 This study resource was shared via CourseH DSC1520/202/1 2. Thus the graph of the function f(x) = −2x 2 + 10x − 8 is: x x x (2 , 5 ; 4 , 5 ) Question 5 1. Initial means t = 0 P = 6 000 1 + 29e−0,4×0 = 6 000 30 = 200. 2. If P = 4 000 then 4 000 = 6 000 1 + 29e−0,4t 1 + 29e −0,4t = 3 2 e −0,4t = 1 58 ln(e −0,4t ) = ln  1 58 −0,4 tlne = ln( 1 58 ) t = ln( 1 58 ) −0,4 = 10, = 10,2 years, rounded to one decimal place. 5 This study source was downloaded by from CourseH on :58:30 GMT -06:00 This study resource was shared via CourseH Question 6 1. log(Q) − log  Q Q+1 = log Q Q Q+1 ! = 0,8 log(Q + 1) = 0,8 Q + 1 = 100,8 Q = 100,8 − 1 Q = 5, = 5,31,rounded to 2 decimal places. 2.  4L 2 L−2 2 = (4L 2 × L 2 ) 2 = (4L 4 ) 2 = 16L 8 . Question 7 1. 3 ln(2x 2 ) − 5 ln x = 7 ln 8x 6 x5 = 7 8x = e 7 x = 1096 8 = 137. 2. log3 12,34 ln 12,34 = ln 12,34 ln 3 × 1 ln 12,34 = 0,9102. Question 8 The cost function is a linear function. We need two points to draw the line of the cost function. Choose the two points where the lines cut through the x-axis (x−axis intercept, thus y = 0) and y-axis (y−axis intercept, x = 0). Calculate (0; y) and (x; 0) and draw a line through the two points. Therefore: If P = 0 then C (P) = 432 000 − 1 800 (0) = 432 000 the point (0 ; 432 000), 6 This study source was downloaded by from CourseH on :58:30 GMT -06:00 This study resource was shared via CourseH DSC1520/202/1 and if C(P) = 0 then 0 = 432 000 − 1 800 (P) 1 800P = 432 000 P = 43200 1800 = 240 the point (240 ; 0). The revenue function is a quadratic function. We need to determine the vertex, roots and y-intercept of the function to draw it. Now given the function R(P) = 6 000P − 30P 2 with the coefficients equal to a = −30; b = 6 000; and c = 0. x = −b 2a = −(6 000) 2(−30) = 100 y = 6 000(100) − (30)(100)2 = 300 000 The roots are the value of the quadratic formula with a = −30; b = 6 000; and c = 0. Therefore x = −6 000 ± p 6 0002 − 4(−30) 2(−30) = −6 000 ± 6 000 −60 = 0 or 200. The graphs of the two functions are: 7 This study source was downloaded by from CourseH on :58:30 GMT -06:00 This study resource was shared via CourseH The break-even points are where the cost function is equal to the revenue function. These are the points A and B. Profit is revenue minus cost therefore the profit area is where the revenue is bigger than the cost (revenue function lies above cost function) and the loss is where the cost is bigger than the revenue (cost function lies above revenue function).