Maximizing Profit _ help please

michaelor

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Mar 7, 2006
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Any help would be greatly appreciated. I have an idea for a formula, but I think I am off base.

Here's what I have:

P = 15(300-20t) + 3(300 - 20t)
I was planning on graphing that when I got hoem and finding the max point, but I have doubts about that method.

Here's the problem:

A fruit grower wishes to ship early in the season when prices are high and spoilage is low. He now has 15 tons on hand and estimates he can add 3 tons per week by waiting, His present profit is $300 per ton; and he estimates that will reduce by $20 per ton, per each week he delays. In how many weeks should he sell to maximize his profit?

Thank you.
 
michaelor said:
P = 15(300-20t) + 3(300 - 20t)
What you've posted is just a straight line, which simplifies as:

. . . . .P = 18(300 - 20t) = -360t + 5400

It's maximum, so called, would occur at your smallest value of t, which would presumably be t = 0.

Please reply showing your steps and reasoning, describing what "t" stands for and how you arrived at your equation.

Thank you.

Eliz.
 
Re

My thinking was, T = times (weeks)
15(300 -20t) + 3(300-20T)

15 tones times 300 per ton, minus 20 dollars per week. Plus 3 tons times 300 per ton - 20 dollars per additional Week.

I know that did not look right, but I have been out of college for a while now, and can not remember the correct procedure. Just trying to help someone out with this, and it's been driving me crazy. If anyone has the correct formula I would appreciate it.
 
The total profit function you need comes from:
Total Profit = (# of Items) x (Profit/ item)
Imagine a table:

Week.. Tons x Profit/ton
0 ........ 15 x 300
1 ........ 18 x 280
2 ........ 21 x 260

etc

Correctly represent the number of tons
in terms of t, the number of weeks, and
the profit / ton sold as a function also of
t. Then just maxime the quadratic function
you obtain for the total profit.
 
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