Monday, December 04, 2006

4.8: Applications to Business and Economics

If any of you used to think that calculus does not have to do with the real world you were mistaken!!! This lesson talks about using calculus concepts and applying them to economics, business, and money- things that can relate to our every day lives!

There are the different functions that were represented in this lesson- cost, marginal cost, average cost, demand (or price), revenue, marginal revenue, profit, and marginal profit. I know all of these sound really similar, like, what's the difference between revenue and marginal revenue? Well, the word marginal should make you think of a derivative. Because the revenue function is represented as R, the marginal revenue function is R'(x) (or the derivative of R). The same thing applies to the rest of the equations as well- ie. if profit is P marginal profit is P'(x) (or the derivative of P).

So here are the actual equations for the money functions:

Cost: C(x)
Marginal cost: C'(x)
Average cost: C(x)/ (x)
Demand: p(x) --- note that this p is lower case!
Revenue: R= xp(x)
Marginal revenue: R'(x)
Profit: P(x) = R(x) - C(x)
Marginal profit: P'(x)

Now that we know all the equations we can start to understand them!

Important things to remember:

-If the average cost is a minimum (if you want to minimize the cost of an item), the marginal cost must equal the average cost, so C'(x) = C(x) / x

-When you're talking about the marginal cost, you're talking about how much $ you are making by selling or producing items.

-The average cost is the total cost in terms of units.

-If the profit is a maximum (you want to maximize the profit), the marginal cost must equal the marginal revenue, so R'=C' and R'-C'=0.

A problem that might be asked:

The cost of producing x amount of ribbon is C(x)= 2000 + 10x + .005x^2.

A. Find the cost (C(x)) the average cost, (c(x)) and the marginal cost (C'(x))of producing 50, 500, and 5000 ribbons?

B. When will the average cost of the ribbons be the lowest?

First you have to find the equations for the cost, av. cost, and marginal cost.

You know the equation for the cost is C(x)=2000 + 10x + .02x^2.
The equation for the average cost is C(x) / x, so when you divide the cost equation by x you get: 2000/x + 10 + .005x. So c(x)= 2000/x + 10 + .005x.
The marginal cost is just the derivative, so C'(x)= 10 + .01x.

Now, because there are so many equations and you have to find different values, plug in 50, 500, and 5000 into each equation. You can even make table to make it easier.

So now we have part A answered! (The answers would be the numbers in the table)

On to part B.

The average cost of the ribbons will be the lowest when the marginal cost= average cost! So this means the equation we found for c(x) = C'(x).

2000/x + 10 + .005x = 10 + .01x

Subtract 10 from both sides.

2000/x + .005x = .01x

Subtract .005x from .01x.

2000/x = .005x

Multiply x to both sides.

.005x^2 = 2000

x= 632.45

Now, plug 632.45 back into the c(x) equation.

2000/ 632.45 + 10 + .005(632.45)

You get 16.32.

So the minimum average cost per ribbon is $16.32.

Now we're done with the question!


This is another question you'll be asked- its like a word problem.

There's a Joshua Radin concert coming up next week at the Troubador. If the ticket price is $12, 2,000 tickets will be sold. But if the price is dropped to $10, 5,000 tickets will be sold. How many tickets should be sold to maximize the revenue?

Find the slope of (2000,12) and (5000,10) and put this into slope intercpet form.

(The slope is 2 / -3000 which simplifies to 1 / -1500)

p- p1 = m(x- x1)

p - 12 = - 1 / 1500 (x - 2000)


p-12= -1/1500x + 4 / 3

Add 12 to both sides.

p = -1/1500x + 40/3

Now that you have your demand function you have to find the revenue function.

The revenue function (R) = xp(x). So multiply the equation we just found by x. You get:

R= -1/1500x^2 + 40/3x

Find the marginal revenue function (take the derivative)

R' = - 1/750 x + 40/3

The maximum revenue is when the marginal revenue is equal to zero.

So, 0= R'= -1/750x + 40/3

-40/3= -1/750x

x= 10000

Now plug 1000 back into the equation that we have for p.

p = -1/1500(10000) + 40/3

p= 20

So to maximize ticket sales the price of the tickets must be $20!

(I can't believe that with a problem I made up the numbers for I got whole numbers for my answer! Yay!)

Here's an amazing link that explains marginal cost. It even talks to you and it really interactive, theres information on the site on fixed and variable costs & I don't think we're learning that, so just skip down to the bottom to the part on marginal cost.

This site is good too:

I know it's a little hard to read so i'll just tell you what it says:
"What matters most in your pricing decision, average or marginal cost?"
And the guy in the car is saying: "Officer, my trip to Chicago only took 2 and 1/2 hours which equates to 60 miles per hour!" But the officer is saying he went 90 miles per hour. You can see this page to see how the speed the car is going relates to the marginal cost/revenue etc.

Gianna you're next!!!


At 7:28 PM, Blogger Kate said...

Tessa, I tried the link to the site that the comic came from and read a little bit of the script, and OMG how strange! Thank God I'm not going into marketing!

At 6:16 PM, Blogger Tessa said...

haha i know, it was pretty confusing. i'm glad that i'm not going into marketing either. calculus however would be a great career, just kidding.


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