Linear Programming Lecture Notes - Penn State Personal Web Server

We have formulated the general maximization problem in Proble 1.5. Suppose that we
are interested in finding a value that minimizes an objective function z(x1, . . . , xn) subject
to certain constraints. Then we can write Problem 1.5 replacing max with min.
Exercise 3. Write the problem from Exercise 1 as a general minimization problem. Add
any appropriate non-negativity constraints. [Hint: You must change max to min.]
An alternative way of dealing with minimization is to transform a minimization problem
into a maximization problem. If we want to minimize z(x1, . . . , xn), we can maximize
−z(x1, . . . , xn). In maximizing the negation of the objective function, we are actually finding
a value that minimizes z(x1, . . . , xn).
Exercise 4. Prove the following statement: Consider Problem 1.5 with the objective
function z(x1, . . . , xn) replaced by −z(x1, . . . , xn). Then the solution to this new problem
minimizes z(x1, . . . , xn) subject to the constraints of Problem 1.5.[Hint: Use the definition of
global maximum and a multiplication by −1. Be careful with the direction of the inequality
when you multiply by −1.]
2. Some Geometry for Optimization
A critical part of optimization theory is understanding the geometry of Euclidean space.
To that end, we’re going to review some critical concepts from Vector Calculus (Math
230/231). I’ll assume that you remember some basic definitions like partial derivative and
Euclidean space. If you need a refresher, you might want to consult [MT03, Ste07].
We’ll denote vectors in R n in boldface. So x ∈ R n is an n-dimensional vector and we
have x = (x1, . . . , xn).
Definition 1.5 (Dot Product). Recall that if x, y ∈ Rn are two n-dimensional vectors,
then the dot product (scalar product) is:
n
(1.7) x · y =
i=1
xiyi
where xi is the i th component of the vector x.
An alternative and useful definition for the dot product is given by the following formula.
Let θ be the angle between the vectors x and y. Then the dot product of x and y may be
alternatively written as:
(1.8) x · y = ||x||||y|| cos θ
This fact can be proved using the law of cosines from trigonometry. As a result, we have
the following small lemma (which is proved as Theorem 1 of [MT03]):
Lemma 1.6. Let x, y ∈ R n . Then the following hold:
(1) The angle between x and y is less than π/2 (i.e., acute) iff x · y > 0.
(2) The angle between x and y is exactly π/2 (i.e., the vectors are orthogonal) iff x·y =
0.
(3) The angle between x and y is greater than π/2 (i.e., obtuse) iff x · y < 0.
Exercise 5. Use the value of the cosine function and the fact that x · y = ||x||||y|| cos θ
to prove the lemma. [Hint: For what values of θ is cos θ > 0.]
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