Jacques I. Mathematics for Economics and Business
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Step 1
The stationary points are the solutions of the simultaneous equations
f
x
(x, y) = 0
f
y
(x, y) = 0
so we need to solve
3x
2
− 3 + y
2
= 0
2xy = 0
There have been many occasions throughout this book when we have solved simultaneous equations. So far
these have been linear. This time, however, we need to solve a pair of non-linear equations. Unfortunately,
there is no standard method for solving such systems. We have to rely on our wits in any particular instance.
The trick here is to begin with the second equation
2xy = 0
The only way that the product of three numbers can be equal to zero is when one or more of the individual
numbers forming the product are zero. We know that 2 ≠ 0, so either x = 0 or y = 0. We investigate these
two possibilities separately.
Case 1: x = 0. Substituting x = 0 into the first equation
3x
2
− 3 + y
2
= 0
gives
−3 + y
2
= 0
that is,
y
2
= 3
There are therefore two possibilities for y to go with x = 0, namely y =−√3 and y =√3. Hence (0, −√3)
and (0, √3) are stationary points.
Case 2: y = 0. Substituting y = 0 into the first equation
3x
2
− 3 + y
2
= 0
gives
3x
2
− 3 = 0
that is,
x
2
= 1
There are therefore two possibilities for x to go with y = 0, namely x =−1 and x = 1. Hence (−1, 0) and
(1, 0) are stationary points.
These two cases indicate that there are precisely four stationary points, (0, −√3), (0, √3), (−1, 0), (1, 0).
Step 2
To classify these points we need to evaluate the second-order partial derivatives
f
xx
= 6x, f
yy
= 2x, f
xy
= 2y
at each point and check the signs of
f
xx
, f
yy
, f
xx
f
yy
− f
2
xy
Partial Differentiation
390
MFE_C05d.qxd 16/12/2005 10:42 Page 390
Point (0, −√3)
f
xx
= 6(0) = 0, f
yy
= 2(0) = 0, f
xy
=−2√3
Hence
f
xx
f
yy
− f
2
xy
= 0(0) − (−2√3)
2
=−12 < 0
and so (0, −√3) is a saddle point.
Point (0, √3)
f
xx
= 6(0) = 0, f
yy
= 2(0) = 0, f
xy
= 2√3
Hence
f
xx
f
yy
− f
2
xy
= 0(0) − (2√3)
2
=−12 < 0
and so (0, √3) is a saddle point.
Point (−1, 0)
f
xx
= 6(−1) =−6, f
yy
= 2(−1) =−2, f
xy
= 2(0) = 0
Hence
f
xx
f
yy
− f
2
xy
= (−6)(−2) − 0
2
= 12 > 0
and so (−1, 0) is not a saddle point. Moreover, since
f
xx
< 0 and f
yy
< 0
we deduce that (−1, 0) is a maximum.
Point (1, 0)
f
xx
= 6(1) = 6, f
yy
= 2(1) = 2, f
xy
= 2(0) = 0
Hence
f
xx
f
yy
− f
2
xy
= 6(2) − 0
2
= 12 > 0
and so (1, 0) is not a saddle point. Moreover, since
f
xx
> 0 and f
yy
> 0
we deduce that (1, 0) is a minimum.
5.4 • Unconstrained optimization
391
Practice Problem
1 Find and classify the stationary points of the function
f(x, y) = x
2
+ 6y − 3y
2
+ 10
We now consider two examples from economics, both involving the maximization of profit.
The first considers the case of a firm producing two different goods, whereas the second
involves a single good sold in two different markets.
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Partial Differentiation
392
Example
A firm is a perfectly competitive producer and sells two goods G1 and G2 at $1000 and $800, respectively,
each. The total cost of producing these goods is given by
TC = 2Q
1
2
+ 2Q
1
Q
2
+ Q
2
2
where Q
1
and Q
2
denote the output levels of G1 and G2, respectively. Find the maximum profit and the
values of Q
1
and Q
2
at which this is achieved.
Solution
The fact that the firm is perfectly competitive tells us that the price of each good is fixed by the market and
does not depend on Q
1
and Q
2
. The actual prices are stated in the question as $1000 and $800. If the firm
sells Q
1
items of G1 priced at $1000 then the revenue is
TR
1
= 1000Q
1
Similarly, if the firm sells Q
2
items of G2 priced at $800 then the revenue is
TR
2
= 800Q
2
The total revenue from the sale of both goods is then
TR = TR
1
+ TR
2
= 1000Q
1
+ 800Q
2
We are given that the total cost is
TC = 2Q
2
1
+ 2Q
1
Q
2
+ Q
2
2
so the profit function is
π=TR − TC
= (1000Q
1
+ 800Q
2
) − (2Q
2
1
+ 2Q
1
Q
2
+ Q
2
2
)
= 1000Q
1
+ 800Q
2
− 2Q
2
1
− 2Q
1
Q
2
− Q
2
2
This is a function of the two variables, Q
1
and Q
2
, that we wish to optimize. The first- and second-order par-
tial derivatives are
= 1000 − 4Q
1
− 2Q
2
= 800 − 2Q
1
− 2Q
2
=−4
=−2
=−2
The two-step strategy then gives the following:
Step 1
At a stationary point
∂
2
π
∂Q
2
2
∂
2
π
∂Q
1
∂Q
2
∂
2
π
∂Q
1
2
∂π
∂Q
2
∂π
∂Q
1
MFE_C05d.qxd 16/12/2005 10:42 Page 392
= 0
= 0
so we need to solve the simultaneous equations
1000 − 4Q
1
− 2Q
2
= 0
800 − 2Q
1
− 2Q
2
= 0
that is,
4Q
1
+ 2Q
2
= 1000 (1)
2Q
1
+ 2Q
2
= 800 (2)
The variable Q
2
can be eliminated by subtracting equation (2) from (1) to get
2Q
1
= 200
and so Q
1
= 100. Substituting this into either equation (1) or (2) gives Q
2
= 300. The profit function there-
fore has one stationary point at (100, 300).
Step 2
To show that the point really is a maximum we need to check that
< 0, < 0, −
2
> 0
at this point. In this example the second-order partial derivatives are all constant. We have
=−4 < 0 ✓
=−2 < 0 ✓
−
2
= (−4)(−2) − (−2)
2
= 4 > 0 ✓
confirming that the firm’s profit is maximized by producing 100 items of G1 and 300 items of G2.
The actual value of this profit is obtained by substituting Q
1
= 100 and Q
2
= 300 into the expression
π=1000Q
1
+ 800Q
2
− 2Q
1
2
− 2Q
1
Q
2
− Q
2
2
to get
π=1000(100) + 800(300) − 2(100)
2
− 2(100)(300) − (300)
2
= $170 000
D
F
∂
2
π
∂Q
1
∂Q
2
A
C
D
F
∂
2
π
∂Q
2
2
A
C
D
F
∂
2
π
∂Q
1
2
A
C
∂
2
π
∂Q
2
2
∂
2
π
∂Q
1
2
D
F
∂
2
π
∂Q
1
∂Q
2
A
C
D
F
∂
2
π
∂Q
2
2
A
C
D
F
∂
2
π
∂Q
1
2
A
C
∂
2
π
∂Q
2
2
∂
2
π
∂Q
1
2
∂π
∂Q
2
∂π
∂Q
1
5.4 • Unconstrained optimization
393
Practice Problem
2 A firm is a monopolistic producer of two goods G1 and G2. The prices are related to quantities Q
1
and
Q
2
according to the demand equations
P
1
= 50 − Q
1
P
2
= 95 − 3Q
2
MFE_C05d.qxd 16/12/2005 10:42 Page 393
If the total cost function is
TC = Q
1
2
+ 3Q
1
Q
2
+ Q
2
2
show that the firm’s profit function is
π=50Q
1
− 2Q
1
2
+ 95Q
2
− 4Q
2
2
− 3Q
1
Q
2
Hence find the values of Q
1
and Q
2
which maximize π and deduce the corresponding prices.
Partial Differentiation
394
Example
A firm is allowed to charge different prices for its domestic and industrial customers. If P
1
and Q
1
denote
the price and demand for the domestic market then the demand equation is
P
1
+ Q
1
= 500
If P
2
and Q
2
denote the price and demand for the industrial market then the demand equation is
2P
2
+ 3Q
2
= 720
The total cost function is
TC = 50 000 + 20Q
where Q = Q
1
+ Q
2
. Determine the firm’s pricing policy that maximizes profit with price discrimination and
calculate the value of the maximum profit.
Solution
The topic of price discrimination has already been discussed in Section 4.7. This particular problem is iden-
tical to the worked example solved in that section using ordinary differentiation. You might like to compare
the details of the two approaches.
Our current aim is to find an expression for profit in terms of Q
1
and Q
2
which can then be optimized
using partial differentiation. For the domestic market the demand equation is
P
1
+ Q
1
= 500
which rearranges as
P
1
= 500 − Q
1
The total revenue function for this market is then
TR
1
= P
1
Q
1
= (500 − Q
1
)Q
1
= 500Q
1
− Q
1
2
For the industrial market the demand equation is
2P
2
+ 3Q
2
= 720
which rearranges as
P
2
= 360 −
3
/2 Q
2
The total revenue function for this market is then
TR
2
= P
2
Q
2
= (360 −
3
/2 Q
2
)Q
2
= 360Q
2
−
3
/2 Q
2
2
The total revenue received from sales in both markets is
TR = TR
1
+ TR
2
= 500Q
1
− Q
2
1
+ 360Q
2
−
3
/2 Q
2
2
MFE_C05d.qxd 16/12/2005 10:42 Page 394
The total cost of producing these goods is given by
TC = 50 000 + 20Q
and, since Q = Q
1
+ Q
2
, we can write this as
TC = 50 000 + 20(Q
1
+ Q
2
)
= 50 000 + 20Q
1
+ 20Q
2
The firm’s profit function is therefore
π=TR − TC
= (500Q
1
− Q
2
1
+ 360Q
2
−
3
/2 Q
2
2
) − (50 000 + 20Q
1
+ 20Q
2
)
= 480Q
1
− Q
2
1
+ 340Q
2
−
3
/
2
Q
2
2
− 50 000
This is a function of the two variables, Q
1
and Q
2
, that we wish to optimize. The first- and second-order par-
tial derivatives are
= 480 − 2Q
1
= 340 − 3Q
2
=−2
= 0
=−3
The two-step strategy gives the following:
Step 1
At a stationary point
= 0
= 0
so we need to solve the simultaneous equations
480 − 2Q
1
= 0
340 − 3Q
2
= 0
These are easily solved because they are ‘uncoupled’. The first equation immediately gives
Q
1
==240
while the second gives
Q
2
=
340
3
480
2
∂π
∂Q
2
∂π
∂Q
1
∂
2
π
∂Q
2
2
∂
2
π
∂Q
1
∂Q
2
∂
2
π
∂Q
1
2
∂π
∂Q
2
∂π
∂Q
1
5.4 • Unconstrained optimization
395
MFE_C05d.qxd 16/12/2005 10:42 Page 395
Step 2
It is easy to check that the conditions for a maximum are satisfied:
=−2 < 0
=−3 < 0
−
2
= (−2)(−3) − 0
2
= 6 > 0
The question actually asks for the optimum prices rather than the quantities. These are found by substituting
Q
1
= 240 and Q
2
=
into the corresponding demand equations. For the domestic market
P
1
= 500 − Q
1
= 500 − 240 = $260
For the industrial market
P
2
= 360 − Q
1
= 360 −=$190
Finally, we substitute the values of Q
1
and Q
2
into the profit function
π=480Q
1
− Q
2
1
+ 340Q
2
−
3
/
2
Q
2
2
− 50 000
to deduce that the maximum profit is $26 866.67.
D
F
340
3
A
C
3
2
3
2
340
3
D
F
∂
2
π
∂Q
1
∂Q
2
A
C
D
F
∂
2
π
∂Q
2
2
A
C
D
F
∂
2
π
∂Q
1
2
A
C
∂
2
π
∂Q
2
2
∂
2
π
∂Q
1
2
Partial Differentiation
396
Practice Problem
3 A firm has the possibility of charging different prices in its domestic and foreign markets. The corres-
ponding demand equations are given by
Q
1
= 300 − P
1
Q
2
= 400 − 2P
2
The total cost function is
TC = 5000 + 100Q
where Q = Q
1
+ Q
2
. Determine the prices that the firm should charge to maximize profit with price dis-
crimination and calculate the value of this profit.
[You have already solved this particular example in Practice Problem 2(a) of Section 4.7.]
Example
A utility function is given by
U = 4x
1
+ 2x
2
− x
1
2
− x
1
2
+ x
1
x
2
(0 ≤ x
1
≤ 5, 0 ≤ x
2
≤ 5)
where x
1
and x
2
denote the number of units of goods 1 and 2 that are consumed.
MAPLE
MFE_C05d.qxd 16/12/2005 10:42 Page 396
(a) Draw a three-dimensional plot of this function and hence estimate the values of x
1
and x
2
at the
stationary point. Is this a maximum, minimum or saddle point?
(b) Find the exact values of x
1
and x
2
at the stationary point using calculus.
Solution
(a) We can name this function utility by typing
>utility:=4*x1+2*x2-x1^2-x2^2+x1*x2;
and then plot the surface by typing
>plot3d(utility,x1=0..5,x2=0..5);
This surface, rotated so that the origin is to the front of the picture, is drawn in Figure 5.13, which shows
that there is just one stationary point. This is clearly a maximum with approximate coordinates (3, 3).
(b) The partial derivatives are worked out by typing
>derivx1:=diff(utility,x1);
which gives
derivx1:=4–2x1+x2
and then typing
>derivx2:=diff(utility,x2);
which gives
derivx1:=2–2x2+x1
5.4 • Unconstrained optimization
397
Figure 5.13
MFE_C05d.qxd 16/12/2005 10:42 Page 397
The stationary point is then found by setting each of these derivatives to zero and solving the resulting
simultaneous equations. This is achieved by typing
>solve({derivx1=0,derivx2=0},{x1,x2});
which generates the response
x
2
= , x
1
=
so we conclude that the utility function has a maximum point when
x
1
= and x
2
=
8
3
10
3
#
$
10
3
8
3
!
@
Partial Differentiation
398
Maximum point A point on a surface which has the highest function value in comparison
with other values in its neighbourhood; at such a point the surface looks like the top of a
mountain.
Minimum point A point on a surface which has the lowest function value in comparison
with other values in its neighbourhood; at such a point the surface looks like the bottom of
a valley or bowl.
Saddle point A stationary point which is neither a maximum or minimum and at which
the surface looks like the middle of a horse’s saddle.
Key Terms
Advice
As usual we conclude this section with some additional problems for you to try. Practice
Problems 4, 5 and 6 are similar to those given in the text. Practice Problems 7, 8 and 9 are
slightly different, so you may prefer to concentrate on these if you find that you do not have
the time to attempt all of them. Practice Problems 10 and 11 provide practice in using Maple.
Practice Problems
4 Find and classify the stationary points of the following functions:
(a)
f(x, y) = x
3
+ y
3
− 3x − 3y (b) f(x, y) = x
3
+ 3xy
2
− 3x
2
− 3y
2
+ 10
5 A firm is a perfectly competitive producer and sells two goods G1 and G2 at $70 and $50, respec-
tively, each. The total cost of producing these goods is given by
TC = Q
1
2
+ Q
1
Q
2
+ Q
2
2
where Q
1
and Q
2
denote the output levels of G1 and G2. Find the maximum profit and the values of
Q
1
and Q
2
at which this is achieved.
6 The demand functions for a firm’s domestic and foreign markets are
P
1
= 50 − 5Q
1
P
2
= 30 − 4Q
2
and the total cost function is
MFE_C05d.qxd 16/12/2005 10:42 Page 398
TC = 10 + 10Q
where Q = Q
1
+ Q
2
. Determine the prices needed to maximize profit with price discrimination and
calculate the value of the maximum profit.
[You have already solved this particular example in Practice Problem 8(a) of Section 4.7.]
7 A firm’s production function is given by
Q = 2L
1/2
+ 3K
1/2
where Q, L and K denote the number of units of output, labour and capital. Labour costs are $2 per
unit, capital costs are $1 per unit and output sells at $8 per unit. Show that the profit function is
π=16L
1/2
+ 24K
1/2
− 2L − K
and hence find the maximum profit and the values of L and K at which it is achieved.
8 An individual’s utility function is given by
U = 260x
1
+ 310x
2
+ 5x
1
x
2
− 10x
1
2
− x
2
2
where x
1
is the amount of leisure measured in hours per week and x
2
is earned income measured in
dollars per week. Find the values of x
1
and x
2
which maximize U. What is the corresponding hourly
rate of pay?
9 A monopolist produces the same product at two factories. The cost functions for each factory are as
follows
TC
1
= 8Q
1
and TC
2
= Q
2
2
The demand function for the good is
P = 100 − 2Q
where Q = Q
1
+ Q
2
. Find the values of Q
1
and Q
2
which maximize profit.
10 (Maple) Draw the surface representing each of the following functions. By rotating the surface,
estimate the x and y values of the stationary points and state whether they are maxima, minima or
saddle points. Use calculus to find the exact location of these points.
(a)
z = x
2
+ y
2
− 2x − 4y + 15 (0 ≤ x ≤ 4, 0 ≤ y ≤ 4)
(b) z = 39 − x
2
− y
2
+ 2y (−2 ≤ x ≤ 2, −2 ≤ y ≤ 2)
(c) z = x
2
− y
2
− 4x + 4y (0 ≤ x ≤ 4, 0 ≤ y ≤ 4)
11 (Maple) A monopolistic producer charges different prices at home and abroad. The demand functions
of the domestic and foreign markets are given by
P
1
+ Q
1
= 100 and P
2
+ 2Q
2
= 80
respectively. The firm’s total cost function is
TC = (Q
1
+ Q
2
)
2
(a) Show that the firm’s profit function is given by
π=100Q
1
+ 80Q
2
− 2Q
2
1
− 3Q
2
2
− 2Q
1
Q
2
Use calculus to show that profit is maximized when Q
1
= 22 and Q
2
= 6, and find the cor-
responding prices.
(b) The foreign country believes that the firm is guilty of dumping because the good sells at a higher
price in the home market, so decides to restrict the sales to a maximum of 2, so that Q
2
≤ 2.
By plotting π in the region 0 ≤ Q
1
≤ 30, 0 ≤ Q
2
≤ 2, explain why the profit is maximized when
Q
2
= 2. Use calculus to find value of Q
1
, and compare the corresponding profit with that of the
free market in part (a).
5.4 • Unconstrained optimization
399
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