Math help desperately needed!
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More Turbo
Joined: Jan 2003
Posts: 554
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From: Boise, ID
Math help desperately needed!
Oh FTE, I come to you in my hour of need. Here's is the problem:
"The profit function (in thousands of dollars) for a company is given by: P(s)=230+20s-(1/2)s^2, where S is the amount (in hundreds of dollars) spent on advertising. Verify the maximum profit algebraically. Use several sentences to help explain your final answer. In your discussion, use transformations to help explain your situation."
Any ideas???
"The profit function (in thousands of dollars) for a company is given by: P(s)=230+20s-(1/2)s^2, where S is the amount (in hundreds of dollars) spent on advertising. Verify the maximum profit algebraically. Use several sentences to help explain your final answer. In your discussion, use transformations to help explain your situation."
Any ideas???
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More Turbo
Joined: Jan 2003
Posts: 554
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From: Boise, ID
Originally Posted by bigredtruckmi
Doing homework are we???
Pchristman,
Thank you for your help!!! I don't know how you went finding an answer of 20 other than trial and error. Using those two numbers I was able to figure out what transformation model to use. It looks like I just need to wrap things up and be done!
Thanks again guys!
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Joined: Oct 2006
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From: Brandon, MS
Use a spreadsheet and calculate P for s= 0 to 100. Use the spreadsheet graphing function to see the curve. Print the graph and attach to your homework.
[edit] I did it real quick but can't paste the graph in here.
[edit] I did it real quick but can't paste the graph in here.
Last edited by SpyFoxx; Apr 9, 2007 at 09:09 PM.
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Post Fiend
Joined: Mar 2002
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From: Bonita Springs FL
Use a spreadsheet and calculate P for s= 0 to 100. Use the spreadsheet to graph it and you can see the curve. Print the graph and attach to your homework.
The hard part is that you have to use algebra. If you are allowed to use calculus, it's a 30 second problem.
It's just a second degree equation, and you have to find the max of the function.
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More Turbo
Joined: Jan 2003
Posts: 554
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From: Boise, ID
Originally Posted by andym
It's just a second degree equation, and you have to find the max of the function.
So I would have a problem that starts out similar to this:
-(1/2)s^2+20s+230
( + ) ( + )
Like that?
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Joined: Oct 2006
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From: Brandon, MS
Originally Posted by andym
So what happens if S=101? You can't pick a range when presented with a problem like this. You have to mathematically determine the answer.
While I will agree with you, all he asked was how to find the maximum S. If 100 didn't give the top of the curve, it is a simples matter of copying down some more rows in a spreadsheet to increase S. So, I gave an alternative solution.
Then in a few sentances he could have said since it is a 2nd degree equation, there would only be one maximum and it is . . . And if you wanted, you could quickly give him a third way by using calculus.
He didn't say that he was limited to algebra (I think?).
Elder User
Joined: Sep 2003
Posts: 952
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From: Charlotte, NC
Basically, take the derivative of the equation which would be P'(s)=20-s, set that equal to 0, there you solve s=20, take that and plug that back into the original equation, and you get P(s)=430.
I miss this stuff. Mechanical Engineer major math minor, here, finishing up my junior year, so i've seen this for a while.
I miss this stuff. Mechanical Engineer major math minor, here, finishing up my junior year, so i've seen this for a while.
Last edited by fordtrucklover94; Apr 9, 2007 at 09:46 PM.
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Joined: Apr 2007
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Perhaps in simpler terms, P= -.5s^2 + 20s =230 is a quadratic.
The max profit will be at the vertex. Simply graph the equation (parabola) and find the value of P at that point. You can solve it algebraically by solving for the value of the vertex.
The max profit will be at the vertex. Simply graph the equation (parabola) and find the value of P at that point. You can solve it algebraically by solving for the value of the vertex.