Optimization: Maximizing Profit

[molly16]

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?

 

[I like Serena]

Welcome to PF, molly16!

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

 

[molly16]

Welcome to PF, molly16!

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?

 

[I like Serena]

I believe your problem statement says "x thousand units" that each sell at a price of "2000 - 5x" per unit.

 

[molly16]

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!

 

[I like Serena]

You're welcome.