SOLUTION: This one has me bogged down. I am attempting to graph and looking at the table. Is this the correct way of looking at this equation? The total profit (�P� in dollars)

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Question 60834This question is from textbook elementry and intermediate algebra
: This one has me bogged down.
I am attempting to graph and looking at the table.
Is this the correct way of looking at this equation?
The total profit (�P� in dollars) for sales of �x� machines is given by P(x) = - 0.2 x^2 + 300x � 200. Find the following:
a. The profit if 500 machines are sold.
b. How many machines must be sold for maximum profit?
This question is from textbook elementry and intermediate algebra

Found 2 solutions by consc198, ikleyn:
Answer by consc198(59) About Me  (Show Source):
You can put this solution on YOUR website!
a) P(x) = - 0.2 x^2 + 300x – 200
Substitute x = 500 machines.
P(500) = -0.2 x 500 ^ 2 + 300 x 500 - 200
P(500) = -0.2 x 250000 + 150000 - 200
P(500) = -50000 + 149800 = 99800
P(500) = $99800
Profit = $99800
b)The more machines are sold the more profit the company makes.
For every one machine that they sell they make $99.8 profit.

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Answer by ikleyn(53804) About Me  (Show Source):
You can put this solution on YOUR website!
.
This question is from textbook elementry and intermediate algebra
This one has me bogged down.
I am attempting to graph and looking at the table.
Is this the correct way of looking at this equation?
The total profit ('P' in dollars) for sales of 'x' machines is given by P(x) = - 0.2x^2 + 300x - 200.
Find the following:
a. The profit if 500 machines are sold.
b. How many machines must be sold for maximum profit?
~~~~~~~~~~~~~~~~~~~~~~~~


        The solution by @consc198 to part  (b)  is incorrect:
        his reasoning related to part  (b)  is  WRONG.

        For the correct solution to part  (b)  see what follows. 


Function P(x) is a quadratic function. It has the leading coefficient negative at x^2.
It means that the plot of this quadratic function is a downward parabola, which has the maximum.


The ordinate of the maximum point (the ordinate of the vertex) is 

    x%5Bmax%5D = -b%2F%282a%29 = -300%2F%282%2A%28-0.2%29%29 = 300%2F0.4 = 3000%2F4 = 750.


Therefore, the maximum profit is

    P9750%29 = -0.2%2A750%5E2+%2B+300%2A750+-+200 = 112300 dollars.


ANSWER.  For maximum profit, the company should sell 750 machines.

Solved correctly and completely.