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Applied Microeconomics in Managerial Decision Making

This text explains how managerial economics applies microeconomic theory and quantitative tools to optimize company objectives, integrating economic theory into decision-making processes.

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Applied Microeconomics in Managerial Decision Making

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  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

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