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Optimization (applications of Differentation) Problem

  • Thread starter Centurion1
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  • #1
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Homework Statement


The demand function for a product is modeled by

p=56e^-0.000012x

Where p is the price per unit (in dollars) and x is the number of units. What price will yield a maximum revenue.


Homework Equations





The Attempt at a Solution



Ok so i tried taking the derivative to start off with. it is obviously an awkward problem. The derivative i found, and this may be wrong is

(6.72 x 10^-4)e^-1.000012x

if that is right, where from here?
 

Answers and Replies

  • #2
hotvette
Homework Helper
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Derivative (of p with respect to x) is right except for sign.

The problem is to maximize something. First thing you need to do is write an expression for what is to be maximized. Is it p or something else? Second step comes from calculus. Remember how to find the min or max of a function? You correctly imply that a derivative is involved, but what about the derivative?
 
Last edited:
  • #3
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Derivative (of p with respect to x) is right except for sign.

The problem is to maximize something. First thing you need to do is write an expression for what is to be maximized. Is it p or something else? Second step comes from calculus. Remember how to find the min or max of a function? You correctly imply that a derivative is involved, but what about the derivative?
__________________
The derivative is used to find the critical numbers right?

as to setting up the equation would this work.

px = r

and then because you cannot have two variables you must take p and substitute it with 56e^-0.000012x.

so your new equation becomes

56e^-.000012x (times) X

???
 
  • #4
hotvette
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Correct, you want to maximize r = 56xe-.000012x. Also correct that derivatives are used to find critical points (but how?). So, what to you need to do with the expression for r to find its maximum? Calculate it's derivative and do what with it?
 
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  • #5
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set it equal to zero and therefore find the critical numbers?
 
  • #6
hotvette
Homework Helper
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Correct! So, set the derivative of r (with respect to x) equal to zero and solve for x. From that you can determine the optimal value for p.

One thing to watch out for. The problem implies that an integer value for x is needed. If that's really true, you'll need to round x up and down to the nearest integer value and check r using both to find out which one produces the largest r.
 

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