E604
Suggested Answers to Problem Set #2
(Problems 2.3, 2.5, 2.6, 2.9)
2.3. The height of a ball t seconds after it is thrown straight up is -1/2gt2 + 40t (where g is
the acceleration due to gravity)
a. If g=32 (as on the earth), when does the ball reach a maximum height? What is that
height?
h
=
-1/2gt2 + 40t
dh/dt =
-gt
+40 = 0
Given g= 32, t = 40/32 = 1.25 seconds
h = 25 feet
b., If g=5.5 (as on the moon) when does the ball reach a maximum height and what is that
height? Can you explain the reasons for the difference between this answer and the
answer for part (a)?
dh/dt =
-gt
+40 = 0
Given g= 5.5, t = 40/5.5 = 7.27 seconds
h = 145.45 feet.
The weaker gravity of the moon damps the upward thrust of the ball more slowly
c. In general, develop an expression for the change in maximum height for a unit change
in g. Explain why this value depends implicitly on the value of g itself.
Via the envelope theorem.
h*
=
-1/2g t*2 + 40(t*)
thus
dh*/dg =
-1/2t*2
Note from the FONC t*(g) = 40/g. Thus dh*/dg = -800/g2. The optimal h depends on the
value of g because t* is a function of g. (Note that you could have alternatively worked
this problem by inserting the relationship between g and optimal t derived from FONC,
inserted this into the objective function and then took the derivative.
2.5 Suppose U(x,y) = 4x2 + 3y2
a. Calculate U/x, U/y
U/x =
U/y =
8x
6y
b. Evaluate these partial derivatives at x=1, y=2
U/x(x=1) =
U/y (y=2) =
8
12
c. Write the total differential for U
dU = U/x dx + U/y dy = 0
8x dx + 6y dy
d. Calculate dy/dx for dU=0 – that is, what is the implied trade-off between x and y
holding U constant?
dy/dx = -8x/6y =
-4x/3y.
The negative sign indicates that getting more x requires forgoing some y and vice versa.
Further the tradeoff is not linear. Starting at any given values for x and y, an increment
of y requires foregoing 4x/3y of x. The size of what is foregone will change with x and y.
However, to asses the rate of tradeoff, we must express the slope in terms of a single
variable (see g below)
e. Show U =16 when x=1, y=2
Inserting into the objective function
U
=
4(1) 2 + 3(2)2 =
16
f. In what ratio must x and y change to hold U constant at 16 for movements away
from x=1, y=2?
dy/dx = -8(1)/6(2)
=
-8/12 =
-2/3
g. More generally, what is the shape of U=16 contour line for this function? What is
the slope of that line?
Solving the objective for y when U=16 yields
y
=
(2/3)(4-x2)1/2
2.5
2
Y
1.5
1
0.5
0
0
0.5
1
1.5
2
X
The slope of that line is
-2x/[3)(4-x2)1/2]
2.6 Suppose that f(x,y) = xy. Find the maximum value for f if x and y are constrained to
sum to 1. Solve this problem in two ways: by substitution and by using the Langragian
mulitipler method.
a. Substitution method
f(x,y) = xy
with substitution
f(x,y) =
x(1-x)
fx
=
1 – 2x
x
=
½
b. Lagrangian method. Write the constraint in implicit function form:
1-x-y = 0
L
=
xy
-
(1-x-y)
Taking first order conditions
 L/x
=y
-

=
0
 L/y
=x
-

=
0
L/
=1
-
x-y
=
0
Solving
x
=
y
2.5
Thus, 1-2x = 0
x =1/2
2.9. One of the most important functions we will encounter in this book is the CobbDouglas function:
y
=
x1 x2
where  and  are positive constants that are each less than one.
a. Show that this function is quasi-concave using a “brute force” method by applying
Equation 2.107
f1
f2
=
=
x1-1 x2
x1 x2-1
f11
f22
f12
=
=
=
(-1)x1-2 x2
(-1)x1 x2-2
()x1-1 x2-1
Thus, by 2.107
f11 f22
2 f12 f1 f2
+
f22 f12
-2

 -1 2
-1
-1
-1


-1
(-1)x1 x2 (x1 x2 ) 2( )x1 x2 x1 x2 x1 x2 +(-1)x1 x2-2(x1-1 x2)2
(-1) 2 x13-2 x23-2
222 x13-2 x23-2
+ (-1)2x13-2 x23-2
2
2
2 2
3-2
3-2
[(-1) + (-1) - 2  ] x1 x2
Notice that with 0<<1 and 0<<1 each of the terms in the final expression are negative. QED
b. Show that the Cobb-Douglas function is quasi-concave by showing that any contour
line of the form y=c (where c is any positive constant) is convex and therefore that the set
of points for which y>c is a convex set.
c
=
x1 x2 implies that
x1
=
Kx2-/
1/
where K = c .
dx1/dx2=
(-/)x2(--)/<0
d2x1/dx22
= [-(--)/2]x2(--2)/ >0
Thus, for a “level curve” defined by a given value c, the slope between x and y is
negative (the first order condition) and it gets negative at a diminishing rate (the second
order condition). The tradeoff between the two variables is convex. Thus the function
itself is quasi-concave.
c. Show that if  +  >1 then the Cobb-Douglas function is not concave (thereby
illustrating that not all quasi-concave functions are concave).
Concavity requires that f11f22 – f122 >0 and f11<0, f22<0 From the second order conditions
in part (a)
f11
=
(-1)x1-2 x2
f22
=
(-1)x1 x2-2
f12
=
()x1-1 x2-1
Thus
f11f22 –f122
=
=
 (-1)(-1) x12-2 x22-2
22 x12-2 x22-2
2-2
2 -2
 (1--) x1 x2 <0 for +>1