Optimization (applications of Differentation) Problem

In summary, the goal is to find the optimal price for a product so that the company earns the most money. To do this, the derivative of the price with respect to quantity is used to find the critical numbers. Once the critical numbers are found, the optimal price can be determined.
  • #1
Centurion1
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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?
 
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  • #2
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?
 
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  • #3
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
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
set it equal to zero and therefore find the critical numbers?
 
  • #6
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.
 

Related to Optimization (applications of Differentation) Problem

1. What is optimization and how does it relate to differentiation?

Optimization is the process of finding the best possible solution to a problem. In the context of differentiation, optimization involves using the derivative to find the maximum or minimum value of a function. This can help us find the most efficient solution to a problem.

2. What are some real-world applications of optimization problems?

Optimization problems can be found in many fields, including engineering, economics, biology, and physics. Some common examples include minimizing cost or maximizing profit in a business, finding the optimal route for a delivery truck, and maximizing crop yield in agriculture.

3. What is the difference between local and global optimization?

Local optimization involves finding the maximum or minimum value of a function within a specific interval. Global optimization, on the other hand, involves finding the maximum or minimum value of a function over its entire domain. This can be more challenging and often requires advanced techniques.

4. How do you know when you have found the optimal solution to an optimization problem?

In order to find the optimal solution, you need to find the point where the derivative of the function is equal to zero. This point is called a critical point and can be either a maximum or a minimum. To determine which one it is, you can use the first or second derivative test.

5. Are there any limitations to using differentiation for optimization problems?

While differentiation is a powerful tool for solving optimization problems, it does have some limitations. It may not always be possible to find an analytical solution, and in some cases, numerical methods may be needed. Additionally, optimization problems with multiple variables can be more complex and may require advanced techniques.

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