Farmer Jones must determine how many acres of corn

| August 30, 2017

Question
On the other hand, the point (x1 ϭ 15, x2 ϭ 70) is not in the feasible region, because even
though x1 ϭ 15 and x2 ϭ 70 satisfy (2), (4), (5), and (6), they fail to satisfy (3): 15 ϩ 70
is not less than or equal to 80. Any point that is not in an LP’s feasible region is said to
be an infeasible point. As another example of an infeasible point, consider (x1 ϭ 40,
x2 ϭ Ϫ20). Although this point satisfies all the constraints and the sign restriction (5), it
is infeasible because it fails to satisfy the sign restriction (6), x2 Ն 0. The feasible region
for the Giapetto problem is the set of possible production plans that Giapetto must consider in searching for the optimal production plan.
DEFINITION s

For a maximization problem, an optimal solution to an LP is a point in the
feasible region with the largest objective function value. Similarly, for a
minimization problem, an optimal solution is a point in the feasible region with
the smallest objective function value. s

Most LPs have only one optimal solution. However, some LPs have no optimal
solution, and some LPs have an infinite number of solutions (these situations are
discussed in Section 3.3). In Section 3.2, we show that the unique optimal solution to
the Giapetto problem is (x1 ϭ 20, x2 ϭ 60). This solution yields an objective function
value of
z ϭ 3×1 ϩ 2×2 ϭ 3(20) ϩ 2(60) ϭ $180
When we say that (x1 ϭ 20, x2 ϭ 60) is the optimal solution to the Giapetto problem, we
are saying that no point in the feasible region has an objective function value that exceeds
180. Giapetto can maximize profit by building 20 soldiers and 60 trains each week. If Giapetto were to produce 20 soldiers and 60 trains each week, the weekly profit would be
$180 less weekly fixed costs. For example, if Giapetto’s only fixed cost were rent of $100
per week, then weekly profit would be 180 Ϫ 100 ϭ $80 per week.

PROBLEMS
Group A
1 Farmer Jones must determine how many acres of corn
and wheat to plant this year. An acre of wheat yields 25
bushels of wheat and requires 10 hours of labor per week.
An acre of corn yields 10 bushels of corn and requires 4
hours of labor per week. All wheat can be sold at $4 a
bushel, and all corn can be sold at $3 a bushel. Seven acres
of land and 40 hours per week of labor are available.
Government regulations require that at least 30 bushels of
corn be produced during the current year. Let x1 ϭ number
of acres of corn planted, and x2 ϭ number of acres of wheat
planted. Using these decision variables, formulate an LP
whose solution will tell Farmer Jones how to maximize the
total revenue from wheat and corn.
2

3

Answer these questions about Problem 1.
a Is (x1 ϭ 2, x2 ϭ 3) in the feasible region?
b Is (x1 ϭ 4, x2 ϭ 3) in the feasible region?
c Is (x1 ϭ 2, x2 ϭ Ϫ1) in the feasible region?
d Is (x1 ϭ 3, x2 ϭ 2) in the feasible region?

produced and x2 ϭ number of bushels of wheat produced,
reformulate Farmer Jones’s LP.
4 Truckco manufactures two types of trucks: 1 and 2.
Each truck must go through the painting shop and assembly
shop. If the painting shop were completely devoted to
painting Type 1 trucks, then 800 per day could be painted;
if the painting shop were completely devoted to painting
Type 2 trucks, then 700 per day could be painted. If the
assembly shop were completely devoted to assembling truck
1 engines, then 1,500 per day could be assembled; if the
assembly shop were completely devoted to assembling truck
2 engines, then 1,200 per day could be assembled. Each
Type 1 truck contributes $300 to profit; each Type 2 truck
contributes $500. Formulate an LP that will maximize
Truckco’s profit.

Group B
5

Why don’t we allow an LP to have Ͻ or Ͼ constraints?

Using the variables x1 ϭ number of bushels of corn

3 . 1 What Is a Linear Programming Problem?

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