# Math Help - Stumped by an optimization problem

1. ## Stumped by an optimization problem

Hi there. I'm working through "Forgotten Calculus" (Barbara Lee Bleau) and am stuck on an optimisation problem. I'll give the question, show my attempt, and then reveal her (stated, not worked) answer. Would appreciate guidance.
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a) the question

The price equation for a new product is
p= 1,600 - 4x
where p = price and x = # of units produced.
The cost function is 40x + 2,000.
Find the maximum profit and the price which needs to be charged to achieve it.
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b) I'm using an approach which has worked well with a number of similar-looking problems.

What's to be maximized? Profit

Profit = revenues - costs
= (units * price) - (units * costs)
= (x (1600-4x)) - (x (40x + 2000))
= (1600x - 4x^2) - (40x^2 + 2000x)
= 1600x - 4x^2 - 40x^2 - 2000x
= -400x - 44x^2

Even though I'm already struggling to see how this product will make anything but losses, I seek out maxima via the first derivative

Profit ' = -400 -88x
0 = -400 - 88x
400 = -88x
-400 = 88x
-4.545454 = x

Plugged back into the original function, it means the more units produced, the more money lost. Making a profit seems impossible as it would mean selling a negative number of units.

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Her answer: "A price of $824 leads to a maximum profit of$148,544". I cannot reach this.

Where am I going wrong?

Ken

2. The cost function is total costs for the units produced. Don't multiply by the units.

Profit = revenues - costs
= (units*price) - costs
= x(1600-4x) - (40x+2000).

However, when I maximize this I still don't arrive at the answer given. I get x = 1560/8 = 195, p = 1600-4x = 820, not 824. Profits are different, too.

3. ## Thanks for trying

I appreciate the response -- I was starting to feel like an orphan...

Good point about the cost function. To be honest she didn't phrase the question that well. Also, in the immediately preceding problem in her book, she had a typo, basing an answer on the notion that when 4x=24, then x= 8. Shame, overall it's a good book but the typos drive me nuts. Perhaps this is one too.

Thanks again,

Ken