We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

TEXT
Unlock Solution Steps
Sign in to
Symbolab
Get full access to all Solution Steps for any math problem
By continuing, you agree to our Terms of Use
and have read our Privacy Policy

Study Guides > Business Calculus

Reading: Applied Optimization, Part I

We have used derivatives to help find the maximums and minimums of some functions given by equations, but it is very unlikely that someone will simply hand you a function and ask you to find its extreme values. More typically, someone will describe a problem and ask your help in maximizing or minimizing something. For example, he or she will say something like the following questions:
  • "What is the largest volume package which the post office will take?"
  • "What is the quickest way to get from here to there?"
  • "What is the least expensive way to accomplish some task?"
In this section, we’ll discuss how to find these extreme values using calculus.

Max/Min Applications

Example

The manager of a garden store wants to build a 600 square foot rectangular enclosure on the store’s parking lot in order to display some equipment. Three sides of the enclosure will be built of redwood fencing, at a cost of $7 per running foot. The fourth side will be built of cement blocks, at a cost of $14 per running foot. Find the dimensions of the least costly such enclosure. The process of finding maxima or minima is called optimization. The function we’re optimizing is called the objective function. The objective function can be recognized by its proximity to “est” words (greatest, least, highest, farthest, most, …) Look at the garden store example; the cost function is the objective function. In many cases, there are two (or more) variables in the problem. In the garden store example again, the length and width of the enclosure are both unknown. If there is an equation that relates the variables we can solve for one of them in terms of the others, and write the objective function as a function of just one variable. Equations that relate the variables in this way are called constraint equations. The constraint equations are always equations, so they will have equals signs. For the garden store, the fixed area relates the length and width of the enclosure. This will give us our constraint equation. Once we have a function of just one variable, we can apply the calculus techniques we’ve just learned to find maxima or minima.

Max-Min Story Problem Technique

  1. Translate the English statement of the problem line by line into a picture (if that applies) and into math. This is often the hardest step!
  2. Identify the objective function. Look for “est” words.
    1. If you seem to have two or more variables, find the constraint equation. Think about the English meaning of the word “constraint,” and remember that the constraint equation will have an = sign.
    2. Solve the constraint equation for one variable and substitute into the objective function. Now you have an equation of one variable.
  3. Use calculus to find the optimum values. (Take derivative, find critical points, test. Don’t forget to check the endpoints!)
  4. Look back at the question to make sure you answered what was asked. Translate your number answer back into English.

Example

The manager of a garden store wants to build a 600 square foot rectangular enclosure on the store’s parking lot in order to display some equipment. Three sides of the enclosure will be built of redwood fencing, at a cost of $7 per running foot. The fourth side will be built of cement blocks, at a cost of $14 per running foot. Find the dimensions of the least costly such enclosure.

Solution

First, translate line by line into math and a picture.
Text
The manager of a garden store wants to build a 600 square foot rectangular enclosure on the store’s parking lot in order to display some equipment. Three sides of the enclosure will be built of redwood fencing, at a cost of $7 per running foot. The fourth side will be built of cement blocks, at a cost of $14 per running foot. Find the dimensions of the least costly such enclosure.
Translation
Let x and y be the dimensions of the enclosure, with y being the length of the side made of blocks. Then Area = A = xy = 600 2x + y costs $7 per foot y costs $14 per foot So, Cost = C = 7(2x + y) + 14y = 14x + 21y Find x and y so that C is minimized.
Figure 1 Figure 1
The objective function is the cost function, and we want to minimize it. As it stands, though, it has two variables, so we need to use the constraint equation. The constraint equation is the fixed area A = xy = 600. Solve A for x to get [latex] x = \frac {600}{y} [/latex], and then substitute into C: [latex-display] C = 14(\frac {600}{y}) + 21y = \frac{8400}{y} + 21y [/latex-display] Now we have a function of just one variable, so we can whack it with calculus (find the critical points, etc.) [latex-display] C\prime = - \frac{8400}{y^2} + 21 [/latex-display] C ′ is undefined for y = 0, and C ′ = 0 when y = 20 or y = −20. Of these three critical numbers, only y = 20 makes sense (is in the domain of the actual function)—remember that y is a length, so it can’t be negative. And y = 0 would mean there was no enclosure at all, so it couldn’t have an area of 600 square feet. Test y = 20: (I chose the second derivative test) [latex] C \prime \prime = \frac{16800}{y^3} > 0 [/latex], so this is a local minimum. Since this is the only critical point in the domain, this must be the global minimum. When y = 20, x = 30. The dimensions of the enclosure that minimize the cost are 20 feet × 30 feet.

“Marginal Revenue = Marginal Cost”

You’ve probably heard before that “profit is maximized when marginal cost and marginal revenue are equal.” Now you can see why people say that! (Even though it’s not completely true.)

General Example

Suppose we want to maximize profit. Now we know what to do—find the profit function, find its critical points, test them, etc. But remember that Profit = Revenue – Cost. So Profit′ = Revenue′ – Cost′. That is, the derivative of the profit function is MR – MC. Now let’s find the critical points—those will be where Profit′ = 0 or is undefined. Profit′ = 0 when MR – MC = 0, or where MR = MC. That’s where the saying comes from! Here’s a more accurate way to express this idea: Profit has critical points when Marginal Revenue and Marginal Cost are equal. In all the cases we’ll see in this class, Profit will be very well behaved, and we won’t have to worry about looking for critical points where Profit’ is undefined. But remember that not all critical points are local max! The places where MR = MC could represent local max, local min, or neither one.

Example

A company sells q ribbon winders per year at $p per ribbon winder. The demand function for ribbon winders is given by: [latex] p = 300 - 0.02q [/latex] The ribbon winders cost $30 apiece to manufacture, plus there are fixed costs of $9000 per year. Find the quantity where profit is maximized.

Solution

We want to maximize profit, but there isn’t a formula for profit showing . . . yet. So let’s make one. We can find a function for Revenue = pq using the demand function for p. [latex-display] R(q)=(300 - 0.02q)q = 300q - 0.02q^2 [/latex-display] We can also find a function for Cost, using the variable cost of $30 per ribbon winder, plus the fixed cost: [latex-display] C(q) = 9000 + 30q [/latex-display] Putting them together, we get a function for Profit: [latex-display] \pi(q) = R(q) - C(q) = (300q - 0.02q^2) - (9000 +30q) = -0.02q^2 + 270q - 9000 [/latex-display] Now we have two choices. We can find the critical points of Profit by taking the derivative of π(q) directly, or we can find MR and MC and set them equal. (Naturally, you’ll get the same answer either way.) I’ll use MR = MC this time. [latex-display] MR = 200 - 0.04q [/latex-display] [latex-display] MC = 30 [/latex-display] [latex-display] 300 - 0.04q = 30 [/latex-display] [latex-display] 270 = 0.04q [/latex-display] [latex-display] q = 6750 [/latex-display] The only critical point is at q = 6750. Now we need to be sure this is a local max and not a local min. In this case, I’ll look to the graph of π(q)—it’s a downward opening parabola, so this must be a local max. And since it’s the only critical point, it must also be the global max. Profit is maximized when they sell 6750 ribbon winders.

Licenses & Attributions

CC licensed content, Shared previously

  • Business Calculus. Provided by: Washington State Colleges Authored by: Dale Hoffman and Shana Calaway. Located at: https://docs.google.com/file/d/0B1lkHWwO61QEM0gwOFhES2N5Tlk/edit. License: CC BY: Attribution.