## 5.2 Applications of Maxima and Minima

(This topic is also in Section 5.2 in Applied Calculus or Section 13.2 of Finite Mathematics and Applied Calculus)

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Prereq: To understand this section, you should know how to locate relative and absolute maxima and minima of a real-valued function of a single variable. Press the "Prev Tutorial" button on the sidebar to go to the tutorial on that subject.

Applied problems in which we have to find the maximum or minimum are sometimes called optimization problems. The key to solving optimization problems is to set up the problems so that is amounts to locating the absolute maximum or minimum of a certain function f.

Sometimes, the given problem requires little or no setting up. here is one we can attempt right off the bat.

Weekly Revenue for the latest best-seller, A River Burns Through It, at OHaganBooks.com, is given by

R = -p3 + 33p2 + 9p       (18 ≤ p ≤ 28)

dollars when the price of the book is p dollars. What price should the company charge to obtain the largest revenue, and what is the largest revenue?

In questions of this kind, the function we are trying to optimize is called the objective function. Thus, the objective function if the function R. The endpoints and stationary points are given by:

 Endpoints: p = R = ; p = R = Stationary Point:(only one in the domain) p = R = ; Round to 2 decimal places

Based on the above information,

 OHaganBooks.com should charge for a maximum revenue of

Sometimes, the objective function may depend on more than one variable, as in the following applied optimization problem (taken from Example 2 in Section 5.2 of Applied Calculus).

Maximize A = xy
subject to
x + 2y = 100,
x ≥ 0, and
y ≥ 0.

The quantity A that we are trying to maximize is called the objective function. The conditions that come after "subject to" are called constraints. Thus, we are trying to maximize some objective function subject to one or more constraints:

 Maximize A = xy           subject to Objective x + 2y = 100 Equality constraint x ≥ 0 Inequality constraint y ≥ 0 Inequality constraint

Q How do we solve optimization problems like this?
A Here's how:

Solving an Applied Optimization Problem

To solve any problem of the above type, go through the following procedure. (The first four steps have already been done for us in the above example.)

In GeneralAbove Example
1.Identify the unknown(s).
These are usually the quantities asked for in the problem.

Here, the unknowns are x and y.
2. Identify the objective function.
This is the quantity you are asked to maximize or minimize.

A = xy
3. Identify the constraint(s).
These can be equations relating variables or inequalities expressing limitations on the values of variables.

x + 2y = 100
x ≥ 0
y ≥ 0
4. State the optimization problem.
This will have the form "Maximize [minimize] the objective function subject to the constraint(s)."

Maximize A = xy
subject to
x + 2y = 100,
x ≥ 0, and
y ≥ 0.
5. Eliminate extra variables.
If the objective function depends on several variables, solve the constraint equations to express all variables in terms of one particular variable. Substitute these expressions into the objective function to rewrite it as a function of a single variable. Substitute the expressions into any inequality constraints to help determine the domain of the objective function.

Solve the constraint equation x+2y = 100 for x to obtain
x = 100-2y.
Substitute this is the objective to obtain
A = (100-2y)y = 100y - 2y2.
Substituting the expression for x into the inequality constraint, we get
100 - 2y ≥ 0, or
y ≤ 50.
Combining this with the ohter inequality (for y) gives
0 ≤ y ≤50.
Thus, the optimization problem reduces to the following:
Maxinize A = 100y - 2y2 subject to
0 ≤ y ≤50.
6. Find the absolute maximum (or minimum) of the objective function.
Use the techniques of the preceding topic.

 A has a maximum when y = The maximum value is A = The corresponding value of x =

The next example is similar to, but a little simpler than Example 5 in Section 5.2 of Applied Calculus or Section 13.2 of Finite Mathematics and Applied Calculus

Your Company manufactures automobile alternators, and production is partially automated through the use of robots. Daily operating costs amount to \$100 per laborer and \$16 per robot. In order to meet production deadline, the company calculates that the numbers of laborers and robots must satisfy the condition

xy = 10,000
where x is the number of laborers and y is the number of robots. Assuming that the company wishes to meet production deadlines at a minimum cost, how many laborers and how many robots should it use?

 1. The unknowns are Select one x and y x only y only x and C y and C C only

 2. The objective function is Select one xy = 10,000 P = xy xy - 10,000 = 0 C = 100x + 16y C = xy - 100x + 16y x >= 0, y >= 0 P = 16,000xy

 3. The constraints are Select one No constraints Not enough information xy = 10,000 xy = 10,000, x >= 0, y >= 0 x >= 0, y >= 0 C = 100x + 16y, x >= 0, y >= 0 C = 100x + 16y, xy = 10,000, x >= 0, y >= 0

 4. The optimation problem states: Select one Maximize xy = 10,000 subject to C = 100x+16y, x >= 0, y >= 0 Maximize C = 100x+16y subject to xy = 10,000, x >= 0, y >= 0 Maximize P = xy subject to C = 100x+16y, x >= 0, y >= 0 Minimze xy = 10,000 and C = 100x+16y subject to x >= 0, y >= 0 Minimize xy = 10,000 subject to C = 100x+16y, x >= 0, y >= 0 Minimize C = 10,000-xy subject to 100x+16y = 0, x >= 0, y >= 0 Minimize C = 100x+16y subject to xy = 10,000, x >= 0, y >= 0

 5. Solving the constraint equation for y gives Select one Cannot be solved Not enough information y = 10,000/x y = 10,000 + x y = 10,000 - x y = 100x /16 y = 100x - 16 y = 100x + 16

 Substituting the expression for y in the objective function gives Select one C = 160,000xy C = 100x + 10,000y C = 160,100x C = 100x + 0.0016x C = 100x + 160,000/x C = 100x + 10,000/(16x)

 6. The derivative of the objective function above is C'(x) = The values of x and y that minimize C are x = y = The minimum cost is C =

You now have several options

• Try some of the questions in the true/false quiz (warning: it covers the whole of Chapter 5) by pressing the button on the sidebar.
• Try some of the on-line review exercises (press the "Review Exercises" button on the sidebar.
• Try some of the exercises on Section 5.2 of Applied Calculus, or Section 13.2 of Finite Mathematics and Applied Calculus

Last Updated: February, 2007