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Intermediate Microeconomics Analysis - Solved Midterm Exam | ECON 173, Exams of Microeconomics

Drake UniversityMicroeconomics
Prof. William M. Boal

Material Type: Exam; Professor: Boal; Class: INTERMED MICROECON ANALYSIS; Subject: Economics; University: Drake University; Term: Fall 2008;

Typology: Exams

Pre 2010

Uploaded on 07/30/2009

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Intermediate Microeconomic Analysis (Econ 173)
Drake University, Fall 2008
William M. Boal
MIDTERM EXAMINATION #1 ANSWER KEY
VERSION A
I. MULTIPLE CHOICE
(1)c. (2)e. (3)a. (4)b. (5)d. (6)a. (7)d. (8)d. (9)b. (10f.
(11)d. (12)f.
II. SHORT ANSWER
(1) a. increase b. by 1.2 units.
(2) a. increase b. by 2 %.
(3) a. increase b. by 5.5 units.
(4) a. decrease b. by 6 units.
(5) a. decrease b. by 3 %.
(6) a. increase b. by 1 %.
(7) a. increase b. by 3.6 %.
(8) 3/4 or 0.75.
III. PROBLEMS
(1) a. dy/dx = -4x - 20.
b. Setting dy/dx = 0 yields x= -5. But this does not satisfy the restriction that “x
cannot be negative." However, this function is a parabola, whose derivative is
negative for all x > -5. Hence its maximum value for non-negative x occurs at
the boundary: x* = 0.
c. f(0) = y* = 15.
(2) (i) y = 0.6 x1 + 0.3 x2 .
y/x1 = 0.6. y/x2 = 3. MRS = 0.5.
(ii) y = 2 x10.6 x20.3 .
y/x1 = 1.2 x1-0.4 x20.3 . y/x2 = 0.6 x10.6 x2-0.7. MRS = x1/(2x2).
(iii) y = 4 ln(x1) + 2 ln(x2) .
y/x1 = 4/x1 . y/x2 = 2/x2. MRS = x1/(2x2).
a. Clearly, functions (ii) and (iii) have exactly the same formula for the MRS.
b. That formula is MRS = x1 / (2x2).
(3) a.
5
2
5
52
1
1
3
2
2
1
1
3
21
1
1
1
x
x
xx
x
xx
y
x
x
y
.
b.
3
5
53
3
2
2
1
2
2
2
2
1
2
2
2
xx
x
xx
y
x
x
y
.
pf3
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Partial preview of the text

Download Intermediate Microeconomics Analysis - Solved Midterm Exam | ECON 173 and more Exams Microeconomics in PDF only on Docsity!

Intermediate Microeconomic Analysis (Econ 173) Drake University, Fall 2008 William M. Boal

MIDTERM EXAMINATION #1 ANSWER KEY

VERSION A

I. MULTIPLE CHOICE

(1)c. (2)e. (3)a. (4)b. (5)d. (6)a. (7)d. (8)d. (9)b. (10f. (11)d. (12)f. II. SHORT ANSWER (1) a. increase b. by 1.2 units. (2) a. increase b. by 2 %. (3) a. increase b. by 5.5 units. (4) a. decrease b. by 6 units. (5) a. decrease b. by 3 %. (6) a. increase b. by 1 %. (7) a. increase b. by 3.6 %. (8) 3/4 or 0.75. III. PROBLEMS (1) a. dy/dx = -4x - 20. b. Setting dy/dx = 0 yields x= -5. But this does not satisfy the restriction that “x cannot be negative." However, this function is a parabola, whose derivative is negative for all x > -5. Hence its maximum value for non-negative x occurs at the boundary: x* = 0. c. f(0) = y* = 15. (2) (i) y = 0.6 x 1 + 0.3 x 2. y/x 1 = 0.6. y/x 2 = 3. MRS = 0.5. (ii) y = 2 x 1 0.6^ x 2 0.^. y/x 1 = 1.2 x 1 -0.4^ x 2 0.3^. y/x 2 = 0.6 x 1 0.6^ x 2 -0.7. MRS = x 1 /(2x 2 ). (iii) y = 4 ln(x 1 ) + 2 ln(x 2 ). y/x 1 = 4/x 1. y/x 2 = 2/x 2. MRS = x 1 /(2x 2 ). a. Clearly, functions (ii) and (iii) have exactly the same formula for the MRS. b. That formula is MRS = x 1 / (2x 2 ). (3) a. ^ ^   5

1 1 3 2 2 1 (^31) 1 2 1 1 1 

x x x x x x x y x x y  (^). b. ^ ^  

2 2 1 2 2 2 2 1 2 2

x x x x x y x x y  (^).

IV. CRITICAL THINKING

(1) Level curves must slope down for this function. This follows either from a verbal argument, or from the formula for the slope of a level curve. Verbal argument: Along a level curve, by definition, the value of y must remain constant. Since the partial derivatives are given to be positive in this problem, both x 1 and x 2 have a positive effect on y. So if x 1 increases, then x 2 must decrease for y to remain constant. Conversely, if x 2 increases, then x 1 must decrease for y to remain constant. So the level curves must slope down. Slope formula: The slope of a level curve, with x 1 on the vertical axis and x 2 on the horizontal axis, is given by

2

y x

y x

 . (This formula can be derived from

the equation for the total derivative, with dy set equal to zero.) Clearly, if both partial derivatives are positive, the slope of the level curve is negative. (See also multiple-choice question (8) and short-answer question (4) on the same topic.) (2) If y is inversely proportional to x, then y = a/x, for some constant a. We can rewrite this as y = a x-1. The elasticity of y with respect to x is therefore negative one (-1). VERSION B I. MULTIPLE CHOICE (1)e. (2)c. (3)c. (4)d. (5)c. (6)d. (7)f. (8)c. (9)c. (10)e. (11)f. (12)b. II. SHORT ANSWER (1) a. increase b. by 1.5 units. (2) a. decrease b. by 3 %. (3) a. decrease b. by 0.1 units. (4) a. decrease b. by 9 units. (5) a. decrease b. by 1 %. (6) a. increase b. by 4.6 %. (7) a. increase b. by 3.6 %. (8) 2/5 or 0.4. III. PROBLEMS (1) a. dy/dx = -2x - 5. b. Setting dy/dx = 0 yields x= -2.5. But this does not satisfy the restriction that “x cannot be negative." However, this function is a parabola, whose derivative is negative for all x > -2.5. Hence its maximum value for non-negative x occurs at the boundary: x* = 0. c. f(0) = y* = 7.

(2) (i) y = 0.2 x 1 + 0.6 x 2. y/x 1 = 0.2. y/x 2 = 0.6. MRS = 3. (ii) y = 5.5 + ln(x 1 ) + 3 ln(x 2 ). y/x 1 = 1/x 1. y/x 2 = 3/x 2. MRS = (3x 1 ) / x 2. (iii) y = x 1 0.2^ x 1 0.^. y/x 1 = 0.2 x 1 -0.8^ x 2 0.6^. y/x 2 = 0.6 x 1 0.2^ x 2 -0.^. MRS = (3x 1 ) / x 2. a. Clearly, functions (ii) and (iii) have exactly the same formula for the MRS. b. That formula is MRS = MRS = (3x 1 ) / x 2. (3) a. ^ ^  2  3  2

1 1 1 2 1 2 1 1 1 

x x x x x x y x x y  (^). b. ^ ^  2  3  3

2 2 1 2 2 1 2 2 2 

x x x x x x y x x y  (^). IV. CRITICAL THINKING Same as Version A. [end of answer key]