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1) Using a graphing utility to zoom in on the maximum point of the revenue function p=40-(5x^2). Use a setting of (1.62 >= x >= 1.65) and (43.5 >= x >= 43.6). Use the trace feature to improve the accuracy of the apporximation obtained in the example. Do you think this improved accuracy is appropiate in the context of this particular problem? Does it change the price?
2) This problem is comparing to a graph of p=40-(5x^2) with points (0,40) and ((sqrt8), 0).

I believe that 1) graph is inappropiate comparing to 2) because 2) deals with maximum revenue more...im not sure how to explain this though. When i graph the two graphs 1) gives me a different trace than 2). I hope that you understand what im trying to say. Need help with this as much as possible. thanks

2007-07-09 20:16:41 · 3 answers · asked by Bigboi924 1 in Science & Mathematics Mathematics

Actualy the graph 1) is comparing to a graph with R=40-(5x^3) for 0 >= x >= (sqrt 8). not p=40-(5x^2). sorry.

2007-07-09 20:42:19 · update #1

3 answers

2) (0,40)

2007-07-14 07:13:35 · answer #1 · answered by Anonymous · 0 0

Not sure what the trace function on your calculator does... from the context of the problem it sounds as if it does some sort of functional approximation over the selected region.

In amy case, in 1) you are looking at very small regions of the graph. This makes the graph look nearly flat, and makes the curvature hard to pick up, so estimating where the maximum will be is difficult. In 2) you are looking at a more representative section of the graph, so your results are likely to be more accurate.

Of course it's obvious from the form of the function that the maximum is at (0, 40), but never mind. ;-)

2007-07-09 20:34:02 · answer #2 · answered by Scarlet Manuka 7 · 0 0

no longer particular what the hint place on your calculator does... from the context of the undertaking curiously it does some variety of smart approximation over the chosen community. In amy case, in a single million) you're gazing at very small aspects of the graph. This makes the graph look on the component to flat, and makes the curvature difficult to be sure upon up, so estimating wherein the utmost may be is hard. In 2) you're gazing at a extra representative area of the graph, so your consequence often are extra terrific. Of course it particularly is seen from the variety of the situation that the utmost is at (0, 40), although in no way ideas. ;-)

2016-12-10 07:29:17 · answer #3 · answered by Anonymous · 0 0

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