Problem 4.16
The problem is to maximize the function
M(s1 , s2 ) =
R(s1 , s2 )
,
C(s1 , s2 )
where R is the revenue from the sales and C the manufacturing costs,
R(s1 , s2 ) = 339s1 + 399s2 − 0.01s21 − 0.007s1s2 − 0.01s22
C(s1 , s2 ) = 400, 000 + 195s1 + 225s2 .
Since maximizing M is equivalent to minimizing −M, we code −M as objective function
in Matlab
function [f,gradf]=objfun(x)
f1=339*x(1)+399*x(2)-0.01*x(1)^2-0.007*x(1)*x(2)-0.01*x(2)^2;
f2=400000+195*x(1)+225*x(2);
f=-f1/f2;
grad1f=((339-0.02*x(1)-0.007*x(2))*f2-195*f1);
grad2f=((399-0.02*x(2)-0.007*x(1))*f2-225*f1);
gradf=[grad1f;grad2f]/f2^2;
The objective function is very flat. By experimenting a while, the region in which the
maximum is located is narrowed down to
3200 ≤ s1 ≤ 3400, 5000 ≤ s2 ≤ 5200.
Executing the commands
x0=[3400;5200];
options = optimset(’GradObj’,’on’);
[xopt,fopt] = fmincon(’objfun’,x0,[],[],[],[],[3200;5000],[3400;5200],[],options)
leads to the following answer
xopt =
1.0e+003 *
3.30366979593808
5.09632706307705
fopt =
-1.21715632659868
1
Let’s check this answer analytically. The necessary conditions for a maximum are
∂M
∂s1
∂M
C2
∂s2
C2
= (339 − 0.02s1 − 0.007s2)C − 195R = 0
= (399 − 0.007s1 − 0.02s2 )C − 225R = 0,
from which it follows that
225(339 − 0.02s1 − 0.007s2 ) = 195(399 − 0.007s1 − 0.02s2).
We can solve this linear equation for s2 ,
s2 = 1.34838709677419s1 + 658.0645161290322,
and then substitute s2 into the first of the two gradient equations to obtain
0.203360625s21 + 4.4726175s1 − 3684501 = 0.
This quadratic equation has one positive and one negative solution. Only the positive
solution is of interest, and after substituting this into the expression for s2 we find the
corresponding value of s2 . The final answer is
s∗1 = 3296.61217535090 ≈ 3297, s∗2 = 5103.17383644089 ≈ 5104,
with M(s∗1 , s∗2 ) = 1.21715661352767. The level contours M = 1.217, 1.2171, 1.21715,
1.217156 shown in the figure below clearly demonstrate that (s∗1 , s∗2 ) is a maximum point.
The prices for this solution point are p∗1 = 290.72, p∗2 = 334.78. Note that the solution
is in the feasible region defined by the inequality constraints of the constrained optimization problem of Example 4.10. Thus there is no need to solve the constrained problem
separately.
5300
5250
s2
5200
1.217156
5150
5100
5050
1.21715
5000
4950
4900
1.2171
1.217
3100 3150 3200 3250 3300 3350 3400 3450 3500
s
2
1