1 / 16

Answers

Answers. BBEM 4103 Managerial Economics Test I. Question 1. Managerial economics can be best described as applied microeconomics because it applies microeconomics theory with quantitative tools to achieve company’s objectives.

kimberly
Download Presentation

Answers

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Answers BBEM 4103 Managerial Economics Test I

  2. Question 1 • Managerial economics can be best described as applied microeconomics because it applies microeconomics theory with quantitative tools to achieve company’s objectives. • It integrates economic theory into the management’s decision making process. • It integrates economics theory with techniques of quantitative analysis such as mathematical economics, optimization analysis, regression analysis, costing, linear programming, and risk analysis.

  3. Question 2 Given the total revenue (TR) and total cost (TC) functions of Firm XY are as follow: TR=1400Q-6Q2 TC=1500+80Q • Write the firm’s profit (π) function. π = TR-TC π = (1400Q-6Q2)-(1500+80Q) π = 1400Q-6Q2-1500-80Q π = 1320Q-6Q2-1500 • Calculate the marginal profit of the firm based on the answer in part (a). π’ = 1320-12Q • Determine the optimum output or Q* that could optimize the firm’s profit. 1320-12Q=0 -12Q=-1320 Q*=110

  4. Question 2 (cont.) • Verify whether the answer in part (c) contribute to the firm’s optimization effort. S.O.C π’ = 1320-12Q π’’ = -12 Yes, Q*=110 contribute to the firm’s optimization effort since (-ve) value indicate maximum. • Calculate the firm’s total cost (TC) Substitute Q*=110 into total cost function. TC=1500+80Q TC=1500+80(110) TC=10,300 • Calculate the firms total revenue (TR) Substitute Q*=110 into total revenue function. TR=1400Q-6Q2 TR=1400(110)-6(110)2 TR=154,000-72600 TR=81,400

  5. Question 2 (cont.) g. Calculate the firm’s total profit. Substitute Q*=110 into the profit function π = 1320Q-6Q2-1500 π = 1320(110)-6(110)2-1500 π = 145,200 - 72,600 - 1500 π = 71,100

  6. Question 3 (a) Maximize the company’s profit subject to the constraint. To maximize the profit, we need to differentiate the profit function π=-60+140Qx+100Qy-10Qx2-8Qy2-6QxQy and solve for Qx* and Qy*. However, the constraint function 20Qx+40Qy=200 put limitation that the total quantity produced by the firm must equal 200 units.

  7. Question 3 (cont.) Step 1: Find Qx from the constraint function. 20Qx+40Qy = 200 20Qx = 200-40Qy Qx = (200-40Qy)/20 Qx = 10 - 2Qy Substitute Qx = 10 - 2Qy into the profit function (which is also known as the objection function of the firm) π=-60+140Qx+100Qy-10Qx2-8Qy2-6QxQy π=-60+140(10 - 2Qy)+100Qy-10(10 - 2Qy)2-8Qy2-6(10 - 2Qy) Qy

  8. Question 3 (cont.) π=-60+140(10 - 2Qy)+100Qy-10(10 - 2Qy)2-8Qy2-6(10 - 2Qy) Qy Expand the new profit function. π=-60+140(10 - 2Qy)+100Qy-10 (10 - 2Qy) (10 - 2Qy) - 8Qy2 - 6(10 - 2Qy) Qy π = -60+1400 - 280Qy+100Qy – 10(100 – 20Qy – 20Qy + 4Qy2) - 8Qy2 – 6(10Qy - 2Qy2) π = -60+1400 - 280Qy+100Qy – 1000 + 200Qy + 200Qy – 40Qy2 - 8Qy2 – 60Qy + 12Qy2 π = 340 + 160Qy – 36Qy2 Differentiate the profit function to maximize the company’s profit subject to the constraint: π’= 160 – 72Qy

  9. Question 3 (cont.) Then, substitute Qy* = 2.22 in the constraint function to obtain Qx* Qx = 10 - 2Qy Qx = 10 – 2(2.22) Qx* = 5.56 units b. What is the optimum quantity of X? c. What is the optimum quantity of Y? Set the first derivative = 0 and solve for the optimum X and Y quantity. π’= 160 – 72Qy 160 - 72Qy = 0 -72Qy = -160 Qy = -160/ -72 Qy* = 2.22 units

  10. Question 3 (cont.) d. Calculate the company’s total profit. Given: π=-60+140Qx+100Qy-10Qx2-8Qy2-6QxQy Substitute Qy* = 2.22 and Qx* = 5.56 into the profit function: π=-60+140(5.56)+100(2.22)-10(5.56)2-8(2.22)2-6(5.56)(2.22) π= -60 + 778.4 + 222 - 309.136 - 39.427 - 74 π= RM 517.837

  11. Question 4 (a) • R = 12.6 + 22W – 4.1X + 16.3Z (b) • R2 = 0.3175 • R2 is the coefficient of determination. It is used to determine how well the regression line fits the data. • It shows that 31% of changes in the dependent variable (R) can be explained by the independent variables W, X, and Z.

  12. Question 4 (cont.) (c) • F-ratio = 4.660 • Another test of overall explanatory power of the regression is the analysis of variance (ANOVA), which uses the F-statistics or F-ratio. • The meaning of F-ratio cannot be simply interpreted based on the given value. • Student needs to use F-distribution table to compare the calculated F values with the critical value from table.

  13. Question 4 (cont.) (d) • The Standard Error (S.E) of Intercept = 8.34, W = 3.61, X = 1.65, and Z = 4.45. • Standard error of estimation is a measure of the dispersion of the data points from the line of best fit (simply known as the regression line) • The smaller the S.E of estimation, the closer the data points are to the regressed line. • In other words, S.E of estimation is used to indicate the accuracy of a regression model.

  14. Question 4 (cont.) (e) • If W, X, and Z are all equal to 0, • R = 12.6 + 22W – 4.1X + 16.3Z • R = 12.6 + 22(0) – 4.1(0) + 16.3(0) • R = 12.6 (f) • If W = 10, X = 5, and Z = 30, R = ? • R = 12.6 + 22(10) – 4.1(5) + 16.3(30) • R = 701.1

  15. Question 5 (a) • Given y = (3x4 + 5)6. Find the derivative using chain rule. • Let y = u6 and u = 3x4 + 5. Then, y’ @ dy/du = 6u5 and u’ @ du/dx = 12x3 • Apply the Chain rule: y’ @ dy/dx = dy/du . du/dx • y’ = 6u5 . 12x3 • y’ = 72x3u5 • @ • y’ = 6(3x4 + 5)5 . 12x3 • y’ = 72x3(3x4 + 5)5

  16. Question 5 (cont.) b) y = (4x2 – 3)(2x5) • y’ = (4x2 – 3)(10x4) + 2x5 (8x) • y’ = 40x6 – 30x4 + 16x6 • y’ = 56x6 – 30x4 c) y = 7x9 (3x2 – 12) • y’ = 7x9 (6x) + (3x2 – 12)(63x8) • y’ = 42x10 + 189x10 – 756x8 • y’ = 231x10 – 756x8

More Related