Calc student stuck on business application: revenue for 1 additional item

Lebombo
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Homework Statement



A company mines low-grade nickel ore. If the company mines x tons of ore, it can sell the ore for the price p = 225 - 0.25x dollers per ton.

How much revenue will the company make by selling an additional 1 ton of ore on top of what is being produced to maximize revenue.

Homework Equations

I made revenue be: R(x) = 225x - 0.25x^{2}

Critical point: [R'(x) = 0] --> [225 - .5x = 0] --> [x = 450]

Setting up a table shows x = 450 is a max value.

The Attempt at a Solution

Now here is where I get stuck. I don't know what the question is asking me to find.Guess 1) Is the question asking for the total revenue realized by the company at the 451st item? If so, I would probably just evaluate f(451) = 50,624.75

Guess 2) Is the question asking for the revenue of the 451st item by itself? If so, I would probably then evaluate f(451) - f(450) = -225.25 (negative revenue)

Guess 3) Or the approximate revenue of the 451st item? Which I assume would be evaluated using the derivative R'(450) = 0Any of these 3 calculations match what the question is asking for?
 
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Yes the question is poorly worded, my guess would be that you are asked to find the revenue made by producing the amount of ore required to maximize profit, and then add to that the revenue from 1 more ton of ore.
 
any help out there on the interwebs?
 
Oops Poopsilon, didn't realize I had a reply already. Thanks.

So if it is the case that you say, wouldn't adding the exact revenue from one additional ton (451st ton) to the total revenue up to 450 tons be the same value as f(451)?

In other words, [f(451) - f(450)]+f(450) = f(451)

I'll ask my professor what he meant exactly. Thanks for your help.
 
Sounds like they are looking for R(n+1) - R(n). With that information, you could generalize something like, "We can make x dollars selling n things, and we make x+y dollars selling n+1 things. The ratio of y/1 might be really small in comparison to the ratio of x/n. That is, it might not be worth producing anything over the maximized amount.

But that is a horrible question. They might just want R(n+1).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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