# Optimization

In this section we will explore the science of
optimization. Suppose that you are trying

to find a pair of numbers with a fixed sum so that the product of the two
numbers is

a maximum. This is an example of an optimization problem. However, optimization

is not limited to finding a maximum. For example, consider the manufacturer who

would like to minimize his costs based on certain criteria. This is another
example of

an optimization problem. As you can see, optimization can encompass finding
either a

maximum or a minimum.

Optimization can be applied to a broad family of different
functions . However, in

this section, we will concentrate on finding the maximums and minimums of
quadratic

functions. There is a large body of real -life applications that can be modeled
by quadratic

functions, so we will find that this is an excellent entry point into the study
of

optimization.

**Finding the Maximum or Minimum of a Quadratic Function**

Consider the quadratic function

f(x) = −x^{2} + 4x + 2.

Let’s complete the square to place this quadratic function
in vertex form . First, factor

out a minus sign.

Take half of the coefficient of x and square , as in
[(1/2)(−4)]^{2} = 4. Add and subtract

this amount to keep the equation balanced .

Factor the perfect square trinomial, combine the constants
at the end, and then redistribute

the minus sign to place the quadratic function in vertex form.

This is a parabola that opens downward, has been shifted 2
units to the right and 6

units upward. This places the vertex of the parabola at (2, 6), as shown in
Figure 1.

Note that the maximum function value (y-value) occurs at the vertex of the
parabola.

A mathematician would say that the function “attains a maximum value of 6 at x

equals 2.”

Note that 6 is greater than or equal to any other y-value
(function value) that occurs

on the parabola. This gives rise to the following definition.

Figure 1. The maximum value of the

function, 6, occurs at the vertex of the

parabola, (2, 6).

**Definition 1**. Let c be in the domain of f. The function f is said to achieve a

maximum at x = c if f(c) ≥ f(x) for all x in the domain of f.

Next, let’s look at a quadratic function that attains a minimum on its domain.

**Example 2. **Find the minimum value of the quadratic
function defined by the

equation

f(x) = 2x^{2} + 12x + 12.

Factor out a 2.

(3)

Take half of the coefficient of x and square, as in
[(1/2)(6)]^{2} = 9. Add and subtract

this amount to keep the equation balanced.

Factor the trinomial and combine the constants, and then
redistribute the 2 in the next

step .

The graph is a parabola that opens upward, shifted 3 units
to the left and 6 units

downward. This places the vertex at (−3,−6), as shown in Figure 2. Note that the

minimum function value (y-value) occurs at the vertex of the parabola. A
mathematician

would say that the function “attains a minimum value of −6 at x equals −3.

Figure 2. The minimum value of the

function, -6, occurs at the vertex of the

parabola, (−3,−6).

Note that −6 is less than or equal to any other y-value
(function value) that occurs on

the parabola.

This last example gives rise to the following definition.

**Definition 4**. Let c be in the domain of f. The function f is said to achieve a

minimum at x = c if f(c) ≤ f(x) for all x in the domain of f.

**A Shortcut for the Vertex**

It should now be clear that the vertex of the parabola
plays a crucial role when optimizing

a quadratic function. We also know that we can complete the square to find the

coordinates of the vertex. However, it would be nice if we had a quicker way of
finding

the coordinates of the vertex. Let’s look at the general quadratic function

y = ax^{2} + bx + c

and complete the square to find the coordinates of the vertex. First, factor out the a.

Take half of the coefficient of x and square, as in
[(1/2)(b/a)]^{2} = [b/(2a)]^{2} = b^{2}/(4a^{2}).

Add and subtract this amount to keep the equation balanced.

Factor the perfect square trinomial and make equivalent
fractions for the constant terms

with a common denominator .

Finally, redistribute that a. Note how multiplying by a
cancels one a in the denominator

of the constant term.

Now, here’s the key idea. The results depend upon the
values of a, b, and c, but it

should be clear that the coordinates of the vertex are

The y-value of the vertex is a bit hard to memorize, but
the x-value of the vertex is

easy to memorize.

Vertex Shortcut. Given the parabola represented by the quadratic function

y = ax^{2} + bx + c,

the x-coordinate of the vertex is given by the formula

.

Let’s test this with the quadratic function given in **Example 2**

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