High School Tasks - Math Problem Example

1. The profits of the firm at three different output levels are the following: Production (units) Profits 40 2800 50 3900 60 4800 a. Find a quadratic function relating production to profits. General form of quadratic equation: y = ax2 +bx + c Substituting the given with production(y) as a function of profits(x): 2800 = 1600a + 40b + c (1) 3900 = 2500a + 50b + c (2) 4800 = 3600a + 60b + c (3) We now have a system of linear equations. Rearranging and forming the matrix, we have: B3_3 = 1600 40 1 2800 2500 50 1 3900 3600 60 1 4800 Solving: B3_3 = 1600 40 1 2800 R2' = R2-R1(1.5625) 1600 40 1 2800 2500 50 1 3900 0 -12.5 -9/16 -475 3600 60 1 4800 3600 60 1 4800 B3_3 = 1600 40 1 2800 R3' = R3-R1(2.25) 1600 40 1 2800 0 -12.5 -9/16 -475 0 -12.5 -9/16 -475 3600 60 1 4800 0 -30 -1.25 -1500 B3_3 = 1600 40 1 2800 R3' = R3-R2(2.4) 1600 40 1 2800 0 -12.5 -9/16 -475 0 -12.5 -9/16 -475 0 -30 -1.25 -1500 0 0 0.1 -360 From R3 we can solve c: 0.1c = -360 or c = -3600 Substituting in R2: -12.5b -9/16c = -475 or b = 200 Substituting in R1: 1600a +40 + c =2800 or a = -1 Hence: y = -x2 + 200x -3600 b. For what range of the firm is profitable and what is the optimum level of output for the firm To determine the range of profitability, we must set y = f(x) = 0 and solve for x: Hence: -x2 + 200x -3600 = 0 By factoring: (-x + 20) (x - 180) = 0 Hence: x = 20 and 180 The range of profitability is therefore: from 20 to 180 units of production. If the production is less than 20 or more than 180, the x values will be negative indicating loss of profit. To find the optimum, the derivative of f(x) should be set to 0: f '(x) = -2x + 200 = 0 Hence: x = 100 c. Graph the function: Profit The given graph is parabolic. As expected, the profit is positive from x = 20 to x =180 verifying our answer in b. It can also be seen that the maximum profit can be found in x = 100. 2. The population of the country is 50 million. Two months ago, the government required its citizen to purchase an identity card. After one month, 6 million people had it and by the end of the second month, 10 million people had one. a. Model N (number of cards) as a function of t (months) using the form: N = a + b/ (t+c) Given are the following: When t = 1 then N = 6,000,000 hence: 6,000,000 = a + b/(1+c) [eq.1] t = 2 N = 10,000,000 hence: 10,000,000 = a + b/(2+c) [eq.2] and of course, when t = 0 N = 0 hence: 0 = a + b/c or a= -b/c [eq.3] Simplifying eq. 1 and inserting the value of a from eq. 3, we have: 6,000,000 = -b/c + b/(1+c) 6,000,000 = (-b -bc +bc) / [c*(1+c)] 6,000,000 = -b/[c*(1+c)] but -b/c = a 6,000,000 = a / (1+c) a = 6,000,000 + 6,000,000c [eq.4] Simplifying eq. 1 and inserting the value of a from eq. 3, we have: 10,000,000 = -b/c + b/ (2+c) 10,000,000 = (-2b -bc +bc) / [c*(2+c)] 10,000,000 = -2b/[c*(2+c)] but -b/c = a 10,000,000 = 2a / (2+c) a = 10,000,000 + 5,000,000c [eq.5] By equation 4 & 5, we can get easily get the value of a & c: a = 30,000,000 c =4 By eq. 3, a = - b/c or b = -ac. Hence: b = -120,000,000 The model equation is therefore: N = 30,000,000 + (-120,000,000)/ (t +4) b. What is the function called Graph and define its features: The function is of the type Rational function because the equations can be expressed as a ratio of two polynomial function. As we can see, the graph is only defined when x is not equal to 4 which is necessary because the denomitor t-4 will be undefined. We can also see that as x approaches positive infinity, the graph tends to flatten indicating a limit. c. How long will it take for 50% of the population to have their identity card Find t (months) when N =25,000,000 25,000,000 = 30,000,000 - 120,000,000/ ( t + 4) -5,000,000 (t+4) = -120,000,000 t + 4 = 24 t = 20 Hence, it will take 20 months for half of the population to get an identity card. d. Price of card is 20 and cost of production and distribution is 10. Find the profit after 4 years. t for 4 years is equal to 48 months Find N when t = 48 N = 30,000,000 - 120,000,000/ ( 48 + 4) N = 27692307.69 Since net profit is (20 - 10) = 10, profit for 4 years is N*10 = 276,923,077. 3. The government plans to generate 250,000 MW of wind power. 3 years ago, 50,000 MW was made available. 2 years ago, an additional 10,000 was added to the grid. A year ago, the total was 70,000 MW. An economist proposed the following equation to model this development: MW = 250,000 - 200,000e-0.05t a. Verify the economist model according to the evidence: To solve this problem, we need only to input t and compute for theoretical MW. Then we compare this to the actual MW. Time (t) Actual MW Theoretical MW Percentage Error (%) 0 50,000 50,000 0 1 50,000+10,000 = 60,000 59754.12 0.41 2 70,000 69,032.52 1.38 Since the percentage error is very small, we can confidently accept the model. b. Plot the graph and identify the function. The equation is an exponential function because it involves the natural number e. c. The government claims that 80% of the MW shall be achieved in 10 years. Verify with the model. MW = 0.80 * 250,000 = 200,000 MW t = time in years Verify: MW = 250,000 - 200,000e-0.05t MW = 250,000 - 200,000e-0.05(10) MW = 128,693.87 From the results, the projection of the government is inaccurate. d. When will they achieve 10% of the target MW = 0.10 * 250,000 = 25,000 t = time in years Determine t: MW = 250,000 - 200,000e-0.05t 25,000 = 250,000 - 200,000e-0.05(t) e-0.05(t) = 9/8 To solve this equation, we need to get the natural logarithm(ln) of both sides) -0.05t ln(e) = ln (9/8) note: ln (e) = 1 -0.05t = ln (9/8) t = -2.36 From the results, the model and even the graph indicate a negative value. Even from the start t = 0, the theoretical MW is already 50,000 telling us that the model is inaccurate in modelling less values below 50,000. 4. Pinewood revenue for 'dining chair' as a function of output was modelled as R = -0.25q2 + 150q Weekly production costs stand at 4096 and unit cost is 70. a. Determine weekly profit function and identify the type of function. Revenue = -0.25q2 + 150q Cost = 4096 + 70q Profit = Revenue - Costs = -0.25q2 + 150q - 4096 - 70q Profit = -0.25q2 + 80q - 4096 The function is quadratic in nature b. Determine range of output and optimum output for profit generation. With the profit equation, we note that a = -0.25, b= 80 and c =-4096. Hence, we can use the quadratic equation for solving the range: q = -b + (b2+4ac) 4ac q = -80 + (802+4(-0.25)(-4096) 4 (-0.25) (-4096) q = 64 & 256 Hence: the range of profitable output is q = 64 to 256 units For optimum value, we must get the derivative of the equation and set it to 0. Profit' (q) = -0.5q + 80 = 0 q = 160 units c. Determine equation for average costs per unit and identify what type of mathematical function. Average cost/ unit = Weekly cost / Unit + actual cost per uit Average cost = 4096 / x + 70x d. Average cost = (4096 + 70x2) / x Rational function e. What happens to average costs when output is very small or output is very large To know what happens, we need to graph the equation: It can be discerned that as the output becomes very small, the average costs becomes very high. This is also true when the output is very high. e. Graph the revenue, cost, profit and average costs against output. From the graphs, it can be seen that the profit graph is entirely the result of subtracting the cost values with the revenue values. The intersections of the revenue and cost lines represent also the value where there is no profit or loss. In addition, it can be seen that costs and average costs intersect at some point. This can be determined with the following solution: Average costs per unit = Total Cost (4096 + 70x2)/x = 4096 + 70x 4096x = 4096 Hence, the output level where average cost equals the total cost when output is only 1 unit. 5. V 6. Market research suggests that potential market for a product is 800,000. At year 1, the market penetration has reached 50% or 400,000. At year 2, the market penetration has reached 75% or 600,000. Using the following model below, answer the following questions: S = a + bect a. Find the model for the sales as a function of time. When t = 0 , S =0 hence: a = -be0 or a =-b since e0=1 (eq.1) When t =1, S =400,000 hence: 400,000 = a + bec Since a = -b, we have 400,000 = -b + bec Simplifying, we have: b = 400,000/ (-1+ ec) (eq.2) When t = 2, S=600,000 hence: 600,000 = a + be2c Since a = -b, we have 600,000 = -b + be2c Since b = 400,000/ (-1+ ec), we have: 600,000 = -400,000 / (-1+ ec) + 400,000*e2c / (-1+ ec) Dividing by 100,00 both sides: 6 = -4 / (-1+ ec) + 4 e2c / (-1+ ec) Combining: 6 = (-4 +4e2c) / (-1 + ec) Hence: -6 + 6ec= -4 + 4e2c Solving: 4e2c - 6ec + 2 = 0 or 2e2c - 3ec + 1= 0 Let ec = x, hence: 2x2 - 3x + 1 = 0 Using quadratic equation with a = 2, b= -3, c =1, we have the following values for x: x = 0.5 & x = 1 Substituting: ec = 0.5 ec = 1 Using natural logarithm: clne = ln 0.5 clne = ln 1 Since ln e = 1 & ln 1=0: c = ln 0.5 c = 0 The value for c =0 is discarded because this will make all ect useless. From eq. 2: b = 400,000/ (-1+ ec) = 400,000/ (-1+ eln 0.5) b = -800,000 From eq. 1: a=-b a = 800,000 Finally, we have the following model: S = 800,000 - 800,000etln0.5 b. Since this involves the special e number, this function can be considered as an exponential function. c. Determine sales at t = 3 years. S = 800,000 - 800,000e3ln0.5 S = 700,000 d. Sketch and discuss properties of the graph: The graph indicates that at t = 0, sales is 0 but as time increases, sales will reach the 800,000 mark. Read More