Problem 3.1: Solution
Many good answers gave been given – and solved:
Various sales, manufacturing and service problems (square vs. round tables, Dell vs.
HP computers, large and small paper, manicure vs. massage, corn vs. wheat, vanilla
ice cream vs. chocolate ice cream, growth of different types of apples, different types
of pastries), spacecraft problems (minimal fuel consumption), a task scheduling problem
(most efficient unloading of packages), filling problems (how many matchbox cars and
dolls in a toy chest, how many chili and jalapeno peppers in a vase), a target practice
problem (how many thin and thick arrows), a rock–transportation problem (how many
trucks of each of two types to purchase), the problem of how many bracelets of each of
two types a girl should make, and study problems (balancing study time vs. work and
sleep time, respectively).
Here is a problem for math instructors:
A math professor has m hard working and n less hard working students in her class. She
spends x + y hours in the semester on a certain topic: x hours are used to develop and
advance the theory, and y hours are used for problem solving and recitation sessions. She
cannot spend more than 15 hours on the topic and needs at least 7 hours to cover the
minimal theory requirement. She knows that if y < 3 the students perform badly in the
exam, and if y > 6 the students get bored. Based on experience from previous courses
she also knows that the average grade in the exam is proportional to mx + ny.
The goal is to maximize
F (x, y) = mx + ny
subject to
x+y
x
y
y
≤
≥
≥
≤
15
7
3
6.
The optimal solution depends on m and n: if m > n the solution point is (x, y) = (12, 3),
and if m < n the solution point is (x, y) = (9, 6).
15
y
F=Fopt if m>n
6
3
F=Fopt if m<n
feasible set
0
0
7
x
15
On the other hand, a bad minded professor would like to minimize the objective
function. In this case the optimal solution would always be (x, y) = (7, 3).