Dieter Balkenborg
BEEM103 –Optimization Techniques for Economists
Departments of Economics
University of Exeter
Homework Week 1 –Solutions
Exercise 1 Bring onto common denominator and simplify
2t t2
2t + 2
5t
t
2t
2
t
2
Solution 1
2t t2
2t + 2
5t
t
2t
2
t
=
2
t (2 t)
2 (t + 1)
3t
t
3t2
2 (t + 1)
=
2
Exercise 2 Bring onto common denominator and simplify
a
x
a
x
Solution 2
a
x
a
x
+
a
y
a
y
a(y x)
xy
a(y+x)
xy
=
=
+
a
y
a
y
a (y x)
xy
xy
y x
=
a (y + x)
y+x
Exercise 3 Simplify
p
p p
3
8 x2 4 y 1=z
p p p
2 3 x y5 z
Solution 3
p
p p
3
8 x2 4 y 1=z
=
p p p
2 3 x y5 z
8 2 1
x3 y 4
2
=
(3a)
2
1
2a
(3a)2
4x
2
1
2
3
1
3
=a
1 2
a
2
y
1
z
1
4
3
10
4
1
2
x
z
3
x
3
y
1
2
5
2
=
(3a)(
=
a3 = 9
Exercise 4 Solve
1
2
1
3
2
1
(3a)
z
2a
1
2
4x y
1) ( 2)
9
4
z
1
=
2 1 a(
1 2 2 3 9 7
aaa = a
2
2
2
9
= 2
x+3
x
9
=a
3
8 2 11
x3 y 4 z
2
z
=
1
3
1
2
1
2
x
1
3
5
2
y
z
p
3
x
4p
4
y9z
2) ( 1)
=a
3
=
1
2
Solution 4
3
(x + 3) (x
2
x 3 x+3
3
2
3)
x 3 x+3
3 (x + 3) 2 (x 3)
3x + 9 2x + 6
x+6
x
9
(x + 3) (x
=
,
3)
= 9,
= 9,
= 9
= 0
=
6
Exercise 5 Solve
1r
K
K
2w
(1
)a + b
1=4
1=4
1
= 4
2
r
w
1
4
K 4 = Q , K 4 = 24
w
rr
1
4
1=2
for K
= Q
for b
= c
Solution 5 We have
K
1r
K
2w
1=2
K
1
+ 14
2
1
= 4
2
r
w
1
4
3
K4
and so
1
24
r
w
1
4
3
3
K =
w
2
r
1
3
4
Q3 =
3
Q , K = 24
4
3
w
r
1
4
4
3
4
Q3 ,
w
2 Q4
r
For the second equation we get
b
= c
c
=
b
(1
(1
)a
)a
1
b =
b
=
(c
(1
)a )
1
1
1
=
(c
(1
)a )
Exercise 6 Expand and simplify
+ q n (1
1 + q + q2 +
q)
Solution 6
=
=
=
=
1 + q + q2 +
1 + q + q2 +
1 + q + q2 +
1 + q q + q2
1 q n+1
+ q n (1 q)
+ qn
1 + q + q2 +
+ qn q
+ qn
q + q2 + q3 +
+ q n+1
q 2 + : : : + q n q n q n+1
2
1
(This simple formula gives the formula for the geometric series:
n
X
qt = 1 + q + q2 +
+ qn =
1
t=0
and, for jqj < 1
1
X
qt =
t=0
q n+1
1 q
1
1
q
because q n+1 approaches zero as n becomes large.)
Exercise 7 Simplify
exp (ln (x))
ln (exp (x))
solve
ln x5=2
0:5 ln x = ln 25
Exercise 8 Find the derivative of
(2x + 1)10
ln 1 x3
Solution 7
d
(2x + 1)10 = 10 (2x + 1)9 2 = 20 (2x + 1)9
dx
d
1
3x2
2
ln 1 x3 =
3x
=
dx
1 x3
1 x3
Exercise 9 Sketch the graph of a function y (x) that has all the following properties:
i) y 0 (x) > 0 when x < 0 and when x > 4
ii) y 0 (x) < 0 when 0 < x < 4
iii) y 00 (x) > 0 when x > 3
iv) y 00 (x) < 0 when x < 3:
Solution 8 The graph of a function with the required properties is given in the …rst
picture below. The next two give the graphs of the …rst and the second derivatives. The
function has a peak at x = 0; a trough at x = 4 and an infection point at x = 3. It
is hence increasing / upward sloping for x < 0 and x > 4 and decreasing / downward
sloping for 0 < x < 4. Correspondingly, the …rst derivative y 0 is positive for x < 0 and
x > 4 and negative for 0 < x < 4, as can also be seen in the second graph. The function
is concave for x < 3 and convex for x > 3. Correspondingly the second derivative y 00 is
negative for x < 3 and positive for x > 3; as can be seen in the second graph.
3
y
80
60
40
20
-2
-1
-20
1
2
3
4
5
1
2
3
4
5
x
-40
-60
-80
y 300
200
100
-2
-1
y
x
400
300
200
100
-2
-1
1
2
3
-100
4
5
x
-200
You were only required to sketch a gtraph as above free hand. I could not resist trying
to …nd explicit function with the required properties, but it took me ages and I used maple
quite a lot. The problem is that a cubic function would not work because a cubic has its
in‡ection point exactly in the middle between the peak and the trough (if the two latter
exist). So, having the critical points nailed at x = 0 and x = 4, the in‡ection point had
to be at x = 2. I then tried to use a polynomial of order 5. I assumed that its derivative
would be of the form
x (x 4) x2 + bx + c
The derivative of the latter
d
x (x
dx
4) x2 + bx + c
= 4x3 + (3b
4
12) x2 + (2c
8b) x
4c
would have to have the factor (x
place. Polynomial division yields
4x3 + (3b
12) x2 + (2c
x 3
3) ; in order to have the in‡ection point in the right
8b) x
4c
= 4x2 + 3bx + 2c + b +
2c + 3b
x 3
and so we must have 2c + 3b = 0: In order to avoid additional in‡ection points I also
wanted the quadratic factor 4x2 + 3bx + 2c + b of the third derivative to have no real
root, which works for b = 1 and c = 32 . (Notice form the third graph above that I only
just avoid the additional roots.) I multiplied x (x 4) x2 x + 32 by two to eliminate a
fraction, expanded and integrated to get my function
2 5
x
5
5 4 11 3
x + x
2
3
6x2 + 15
on which the above graphs are based. The constant 15 was only added to make the graph
look clearer, not going through on of the numbers at the coordinate axis etc.
Exercise 10 Consider the function
y (x) = e
x2
i) Calculate and draw a sign diagram for the …rst derivative. Where is the function
increasing or decreasing. Are there any peaks or troughs? Does the function have an
(absolute) maximum.
ii) Calculate and draw a sign diagram for the second derivative. Where is the function
convex or concave. Are there any in‡ection points?
Solution 9
y
1.0
0.8
0.6
0.4
0.2
-5
-4
-3
-2
-1
0
2
1
2
3
4
5
x
a) y 0 = 2xe x . Thus the function has a unique critical point at zero. From the sign of
the derivative we see that the function is increasing to the left and decreasing to the right.
Hence x = 0 is the (global) maximum of the function. Let g (u) = eu and h (x) = x2 .
g (u) is increasing and h (u) is concave. As a monotone transformation of a concave
2
function the function y (x) = e x = g (h (x)) is quasi concave.
2
b) y 00 = 2 ( 1 + 2x2 ) e x . The second derivative is zero at x = p12 ; in between it is
negative, outside positive. x = p12 are hence in‡ection points with the function concave
in between the roots and convex outside.
5
Exercise 11 Find all critical points of the function
1
y (x) = x4
4
2x2
Are these local maxima or minima?
Solution 10
dy
= x3 4x = x x2 4 = x (x + 2) (x 2)
dx
The function has critical points at x = 0; +2; 2.
r !
r !
d2 y
4
4
x
= 3x2 4 = 3 x +
2
dx
3
3
p
p
The function
has
in‡
ection
points
at
x
=
4=3
and
at
x
=
4=3 < 2. It is convex for
p
p
4=3 and x > 4=3. It is concave in between. Hence x = +2 and x = 2 are
x<
local minima while x = 0 is a local maximum. All this is visible from the graph of the
function and its two derivatives:
y
8
6
4
2
-3
-2
-1
1
2
3
1
2
3
1
2
3
-2
x
-4
y
20
10
-3
-2
-1
-10
x
-20
y 25
20
15
10
5
-3
-2
-1
6
x
Exercise 12 For the function
1
y = x4 2x2
4
…nd the maximum a) on the interval [ 2; 2] and b) on the interval [ 4; 4] :
Solution 11 We have
y ( 4) = 32
y ( 2) =
4
y (0) = 0
On the interval [ 2; 2] the minima are hence at 2 and the maximum is at x = 0. On
the interval [ 4; 4] the minima are hence at 2 and the maxima are at x = 0.
Exercise 13 A producer operating in a perfectly competitive market has the total cost
function
T C (Q) = 2Q3 18Q2 + 60Q + 50
where costs are given in Pounds Sterling.
1. Calculate and sketch the marginal cost function M C (Q) and the average variable
cost function AV C (Q).
2. Solve the equation M C (Q) = AV C (Q).
3. At what quantity are average variable costs minimized?
4. What are the minimum average variable costs?
5. What quantity maximizes pro…ts when the market price is P = 10?
6. What is the maximal pro…t the …rm can make at this price?
7. What quantity maximizes pro…ts when the market price is P = 25?
Solution 12 1.
M C (Q) = 6Q2
AV C (Q) = 2Q2
36Q + 60
18Q + 60
y 60
40
20
0
0
1
2
7
3
4
5
x
AVC are drawn fat.
2)
M C (Q) = 6Q2 36Q + 60 = 2Q2
4Q2 18Q = 4Q (Q 4:5) , Q = 0
18Q + 60 = AV C (Q) ,
or
Q = 4:5
3) The minimum of AV C is at 4.5, where AV C and M C meet (in the positive), i.e.
at Q = 4:5. This can also be proved directly:
dAV C
= 4Q 18 = 0 , Q = 4:5
dQ
d2 AV C
= 4 > 0 so the function is convex
dQ2
Minimal average costs are
AV C (4:5) = 19:5
4) At the price P = 10 minimal average costs cannot be covered at any quantity.
Therefore zero output is optimal. The pro…t function is in the case
P
Q
2Q3
T C (Q) = 10Q
18Q2 + 60Q + 50
and has the following graph
0
y
2
4
6
x
-50
-100
-150
-200
5) At best the …rm loses its …xed cost, i.e., the optimal pro…t is 50.
6) Now the minimal average costs are covered by the price and it is optimal to produce
where price equals marginal costs. This gives the equation
M C (Q) = 6Q2
36Q + 60 = 25
p
The solutions to this quadratic equation are Q = 3 + 61 114
4:7795 and Q = 3
p
1
114 1:2205. We have to take the larger root. The pro…t function is in the case
6
P
Q
2Q3
T C (Q) = 25Q
18Q2 + 60Q + 50
and has the following graph
0
y
2
4
-40
-60
-80
-100
8
6
x
Solution 13
ln (exp (x)) = eln x
exp (ln (x))
We have
ln x5=2
0:5 ln x =
5
ln x
2
ln ex = x
0:5 ln x = 2 ln x
and ln 25 = ln 52 = 2 ln 5. Hence
2 ln x = 2 ln 5
and so x = 5.
9
x=0