# Math Help - Maximum profit.

1. ## Maximum profit.

Phillip, the owner of a vineyard, estimates the first 9800 bottles of wine produced this season will fetch a profit of $5 per bottle. Howeer, the profit from each bottle beyond 9800 drops by$0.0003 for additional bottle sold.

Assuming at least 9800 bottle of wine are produced and sold, what is the maximum profit?

What would be the profit per bottle in this case?

2. The profit up to 9800 is $49000. Beyond that now, the profit drops by$0.0003 for each additional bottle.
So, for first bottle (after 9800), the profit is $4.9997 For the second (after 9800), the profit is$4.9994
etc.

The profit of the nth bottle (after 9800) is thus given by $Profit = 5 - 0.0003n$

The sum of profit for the bottles beyond 9800 is then given by:

$S_n = \dfrac{n}{2}\left(a + l\right)$

$S_n = \dfrac{n}{2}\left(4.9997 + 5-0.0003n\right)$

$S_n = \dfrac{n}{2}\left(9.9997 -0.0003n\right)$

$S_n = 4.99985n -0.00015n^2$

So, the total profit becomes:

$Profit_{net} = 49000 + 4.99985n -0.00015n^2$

Now, to find the maximum, we find the derivative and set to zero.

$\dfrac{dP}{dn} = 4.99985 - 0.0003n$

$4.99985 - 0.0003n = 0$

$n = 16666.16$

Well, since it's not an integer let's test both the lower and upper integers.

When n = 16666, P = $90664.1667 When n = 16667, P =$90664.1666

Uh... well... the difference is so small that it is found to a hundredth of a cent...

But you got your answer. I derived the profit formulae from the arithmetric progressions formulae and modified them to suit the situation.

3. Thank you for the reply!!

So the answers are $4.99 and$90664.16?

4. Well, I assume that we are looking for the average profit per bottle, if that's so, then we need to have:

$\dfrac{\text{Total Profit}}{\text{Total bottles}} = \dfrac{90664.1667}{9800 + 16666} = 3.43$

To the nearest cent.

5. They both came out to be wrong.

6. Hm... maybe I treated the problem in the wrong way... does the overall profit per bottle of ALL bottles decrease altogether when 9800 is exceeded?

If so, then it's easier... I think.