Econ 299 Tutor Questions
University of Alberta
Econ 299 Tutor Questions
Chapters 2-4
1) Calculate the following using log and non-log growth formulas:
a) My mark in economics last year was 86%. This year it was 91%.
Calculate growth.
growth 
( X t  X t 1 )
91  86
100 
100  5.81%
X t 1
86
growth  {ln( X t )  ln( X t 1 )}100
growth  {ln( 91)  ln( 86)}100
growth  5.65%
b) On my first assignment, I got 71/100. On my second assignment,
I got 42/50. Calculate growth.
growth 
( X t  X t 1 )
42 / 50  71 / 100
100 
100  15.5%
X t 1
71 / 100
growth  {ln( X t )  ln( X t 1 )}100
growth  {ln( 42 / 50)  ln( 71 / 100)}100
growth  14.4%
2) Calculate dy/dx. Calculate elasticity at x=0 and x=10:
a) y=12+72x
dy
 72
dx
  dy dx x y  72 x /(12  72 x)
 (0)  72(0) /(12  72(0))  0
 (10)  72(10) /(12  72(10))  720 / 732  0.983
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Econ 299 Tutor Questions
b) ln(y)=-5ln(x)-10
dy
dx   5
y
x
dy  5 y / x
dx
  dy dx x y  5( y / x)( x / y )  5
 ( 0 )  5
 (10)  5
c) y=(x-4)2
dy
dx  2( x  4)
y
2( x  4) x
2x
  dy dx x y 

2
x4
( x  4)
2(0)
 ( 0) 
0
04
2(10)
 (10) 
 3.33
10  4
3) Dr. Derivative recently conducted a study on University student
consumption and estimated the following formula:
Consumption = 100 +0.5Income-0.1Income2-0.2Age*Grades
a) Find the marginal propensity to consume and explain its
significance
Mpc=dConsumption/dIncome
=0.5-0.2Income
-a person is willing to spend at most half of each additional dollar earned
-this decreases as the person earns more
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Econ 299 Tutor Questions
b) Calculate age’s effect on consumption, and explain its
significance
dConsumption/dAge =-0.2Grades
-as a person ages, he or she spends less on consumption, all else held
equal
-this effect is greater given higher grades
c) Calculate grades’ effect on consumption, and explain its
significance
dConsumption/dGrades=-0.2Age
-a person with better grades will consume less, all else held equal
-this is effect is greater as one ages
4) Graph the following demand equation. Show all steps:
Q = 2000+10P-P2
i) q(0) = 2000+10(0)-02 = 2000; y-intercept is at 2000
q(∞)≈ 2000+10(∞)-∞2 ≈ -∞; graph ends in the lower right quadrant
q(-∞)≈ 2000+10(-∞)-(-∞)2 ≈ -∞; graph starts in the lower left quadrant
ii) 0=2000+10-p2
p=-b±(b2-4ac)1/2 /2a
=-10±(102-4(-1)2000)1/2 /2(-1)
=-10±(8100)1/2 /-2
=(-10±90)/-2
=-50, 40;
x-intercepts are at -50 and 40
iii) q’=10-2p
0=10-2p
2p=10
p=5
q’(0) = 10
q’(10) = -10
Slope is positive below x=5 and negative above x=5.
iv) q(5) = 2000-10(5)-52
= 2025; (5,2025) is a possible max. or min.
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Econ 299 Tutor Questions
v) q’’ = -2; graph is always concave; (5,2025) is a maximum
vi) graph is always concave
vii) q’’ never equals zero; there are no inflection points
5) Interpret the following equation, not including elasticity:
Sanity = 50+7Marriage-12Kids2
Sanity = a measure of sanity
Marriage = years of marriage
Kids = number of kids
Intercept = 50; an unmarried person with no kids has a sanity of 50.
dSanity/dMarriage = 7; each year of marriage increases a person’s sanity
by 7
dSanity/dKids = -24Kids; having kids decreases a person’s sanity by 24
times the number of kids
d2Sanity/dKids2 = -24; each additional kid increases kids’ negative impact
on sanity by 24
6) Indicate whether the following statements are true, false, or uncertain.
Justify your answers:
a) If the second derivative is equal to zero, an inflection point occurs
at this point.
UNCERTAIN
-Generally, an inflection point occurs when the second derivative is equal
to zero
-if the graph is a straight line or discontinuous, an inflection point does not
occur
b) Given the function f(x,y), Young’s theorem states that the second
derivative with respect to x is equivalent to the second derivative
with respect to y.
FALSE
-Young’s theorem states that fxy=fyx
-This is not equivalent to fxx=fyy
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Econ 299 Tutor Questions
c) A concave function will always have a local maximum.
FALSE
-in a concave function, slope is decreasing
-slope may never reach zero, where a maximum may occur
Example: y=ln(x)
y’=1/x (slope never equals zero)
y’’=-1/x2 (concave function)
7) Find the first and second derivatives of the following:
a) y=xsin(12x2) (6 marks)
b) y=ln(12x)cos(12/x) (6 marks)
a) y’
y’’
b) y’
y’’
= sin(12x2)+xcos(12x2)24x
= sin(12x2)+24x2cos(12x2)
= cos(12x2)24x+[48xcos(12x2)-24x2sin(12x2)24x]
= 24xcos(12x2)+48xcos(12x2)-576x3sin(12x2)
= 72xcos(12x2)-576x3sin(12x2)
=12[cos(12/x)]/12x-ln(12x)sin(12/x)(-12/x2)
=[cos(12/x)] / x + [12 ln(12x)sin(12/x)]/x2
=[-sin(12/x)(-12/x2)x-cos(12/x)] / x2
+[ {12(12/12x)sin(12/x)+12ln(12x)cos(12/x)(-12/x2)}x2
-(2x)12 ln(12x)sin(12/x)]/x4
8) Solve the following by both internalizing the constraint and using the
Lagrangean:
Max y=(x+2)(z-3) subject to the constraint x+z=10
Internalizing the Constraint:
z=10-x
y=(x+2)(10-x-3)=(x+2)(7-x)
y=-x2+5x+14
FOC:
dy/dx=-2x+5=0
5=2x
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Econ 299 Tutor Questions
2.5=x
z=10-x
z=10-2.5
z=7.5
SOC:
d2y/dx2=-2, maximum confirmed
Evaluate:
y=(x+2)(z-3)
y=(2.5+2)(7.5-3)
y=20.25
y is maximized at 20.25 when x is 2.5 and z is 7.5
Lagrangean:
L=(x+2)(z-3)+λ(10-x-z)
FOC:
dL/dx=z-3-λ=0
λ=z-3
(A)
dL/dz=x+2-λ=0
λ=x+2
(B)
dL/dλ=10-x-z=0
10=x+z
(C)
(A+B)
λ=λ
z-3=x+2
z=x+5
(D)
(C+D)
10=x+z
10=x+x+5
5=2x
2.5=x
z=x+5
z=2.5+5
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Econ 299 Tutor Questions
z=7.5
SOC:
g 2  2 y g 2  2 y
g g
SOC  ( )
( )
2
2
2
x z
z x
x x
SOC  12 0  12 0  2(1)(1)  2  0
Maximum!
Evaluate:
Same as above
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