Optimization: Maximizing Profit

In summary, optimization is the process of finding the best solution to a problem by maximizing or minimizing a specific objective function. The objective function in maximizing profit is the mathematical representation of the relationship between inputs and outputs in a business, typically expressed as revenue minus costs. Optimization can be applied in various ways to maximize profit in a business, such as identifying the optimal pricing strategy, determining the most efficient production process, or optimizing the allocation of resources. The key factors to consider when optimizing for profit include demand, production costs, market competition, and resource constraints. The potential benefits of optimizing for profit include increased revenue, reduced costs, improved efficiency, and a competitive advantage in the market.
  • #1
molly16
16
0

Homework Statement



Through market research, a computer manufacturer found that x thousand
units of its new laptop will sell at a price of 2000 - 5x dollars per unit.
The cost, C, in dollars of producing this many units is
C(x) = 15 000 000 +1 800 000x + 75x^2. Determine the level of sales
that will maximize profit.

Homework Equations


Profit = Revenue - Cost

The Attempt at a Solution



I said Revenue = 2000 - 5x and Cost = 15 000 000 +1 800 000x + 75x^2
Using the formula Profit = Revenue - Cost I subtracted them from each other and got:

Profit = -14998000 - 1800005x - 75x^2

Then I found the derivative which came out to be:

P'(x) = -1800005 - 150x

Then I set it equal to zero and solved for x:

0 = -1800005 - 150x
1800005 = -150x
x = -12000.03

but the answer in the back of the book is 19 704 units.
Can anyone explain what I did wrong?
 
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  • #2
Welcome to PF, molly16! :smile:

Your revenue looks to be the price of only 1 unit.
Shouldn't more units be sold?
 
  • #3
I like Serena said:
Welcome to PF, molly16! :smile:

Your revenue looks to be the price of only 1 unit.
Shouldn't more units be sold?

So does that mean I should multiply 2000 - 5x by x to represent the number of units?
 
  • #4
I believe your problem statement says "x thousand units" that each sell at a price of "2000 - 5x" per unit.
 
  • #5
I like Serena said:
I believe your problem statement says "x thousand units" that each sell at a price of "2000 - 5x" per unit.

So:
R = (1000x)(2000 - 5x)

then

Profit = Revenue - Cost
P = (1000x)(2000 - 5x) - 15 000 000 +1 800 000x + 75x^2

and

P' = 200000 - 10150x
x= 19.704

#of units = (1000x) = (1000)(19.704) = 19704 units

which was the answer in the back of the book

Thanks for the help!
 
  • #6
You're welcome. :smile:
 

FAQ: Optimization: Maximizing Profit

What is optimization?

Optimization is the process of finding the best solution to a problem by maximizing or minimizing a specific objective function.

What is the objective function in maximizing profit?

The objective function in maximizing profit is the mathematical representation of the relationship between inputs and outputs in a business, typically expressed as revenue minus costs.

How can optimization be applied to maximize profit in a business?

Optimization can be applied in various ways to maximize profit in a business, such as identifying the optimal pricing strategy, determining the most efficient production process, or optimizing the allocation of resources.

What are the key factors to consider when optimizing for profit?

The key factors to consider when optimizing for profit include demand, production costs, market competition, and resource constraints.

What are the potential benefits of optimizing for profit?

The potential benefits of optimizing for profit include increased revenue, reduced costs, improved efficiency, and a competitive advantage in the market.

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