# Determining the maximum profit given a function

• May 7th 2011, 03:18 PM
Arise01
Determining the maximum profit given a function
So with P(x) = -5x^2 + 550x - 5000 ...

... I'm going to sub in 55 into x to determine the maximum profit. (This is just based off of a question that I've been working on)

P(55) = -5(55)^2 + 550(55) - 5000

I've subbed in everything correctly now I just need to solve it. This is what I've come up with.

P(55) = -5(55)^2 + 550(55) - 5000
0= -5(55)^2 + 550(55) - 5000 - 55
0= -5(55)^2 + 550(55) - 4945

I'm not sure whether I'm doing it right.
• May 7th 2011, 03:31 PM
topsquark
Quote:

Originally Posted by Arise01
So with P(x) = -5x^2 + 550x - 5000 ...

... I'm going to sub in 55 into x to determine the maximum profit. (This is just based off of a question that I've been working on)

P(55) = -5(55)^2 + 550(55) - 5000

I've subbed in everything correctly now I just need to solve it. This is what I've come up with.

P(55) = -5(55)^2 + 550(55) - 5000
0= -5(55)^2 + 550(55) - 5000 - 55
0= -5(55)^2 + 550(55) - 4945

I'm not sure whether I'm doing it right.

You can certainly calculate P(55), but where did you get 55 from? Your profit function is a parabola. Since the leading coefficient is negative we know that the parabola has a maximum. (ie it opens downward.) We also know that the maximum value of a parabola is reached when x is on the axis of symmetry of the parabola. How do you find the axis of symmetry?

-Dan

Edit: By the way. You seem to be acting under the assumption that P(55) = 55. It is not. Use a calculator if you are not convinced.
• May 7th 2011, 03:38 PM
topsquark
Quote:

Originally Posted by topsquark
You can certainly calculate P(55), but where did you get 55 from? Your profit function is a parabola. Since the leading coefficient is negative we know that the parabola has a maximum. (ie it opens downward.) We also know that the maximum value of a parabola is reached when x is on the axis of symmetry of the parabola. How do you find the axis of symmetry?

-Dan

Edit: By the way. You seem to be acting under the assumption that P(55) = 55. It is not. Use a calculator if you are not convinced.

Sorry, I did the last post without actually calculating where the maximum was. Yes, it is at x = 55. (You confused me with those last three lines.)

I'm a little lost. You have that $P(55) = -5(55^2) + 550(55) - 5000$. This is just a number. Why don't you simply calculate it?

-Dan
• May 7th 2011, 03:42 PM
Arise01
The x - intercepts for -5x^2 +550x - 5000 are 10 and 100.

10 + 100 = 110 / 2 = 55.

This thread is actually sort of a continuation to another thread I posted not too long ago. I'm pretty sure 55 is the number I want to sub in. Let's just assume that it is.
• May 7th 2011, 03:45 PM
Arise01
55 = -5(55)^2 + 550(55) - 5000

I'm not sure how to calculate it out. Where do I start?
• May 7th 2011, 03:59 PM
topsquark
Quote:

Originally Posted by Arise01
55 = -5(55)^2 + 550(55) - 5000

I'm not sure how to calculate it out. Where do I start?

Simple arithmetic: $55^2 = 55 \cdot 55 = 3025$ in case you've forgotten how to do squares.

Seriously if you don't know how to do this problem (calculating P(55)) you should not be playing around with quadratics. Please do some review.

-Dan
• May 7th 2011, 04:08 PM
VincentP
You calculated that the maximum is at x=55. To calculate the profit you must find P(55) = -5(55)^2 + 550(55) - 5000 = -5(3025) + 30250 - 5000 = -15125 + 25250 = 10125. Remember P(55) is not the same as 55, it's the Profit when x=55. And you are not trying to solve an equation here you are just evaluating a function at a given point, in this case x=55.
• May 7th 2011, 05:36 PM
Arise01
@ Vincent - Yes I'm aware that x = 55. It's what you sub in to x in order to find out the maximum profit.

@ topsquark - Math is just a difficult subject for me to grasp. I'm doing my best to try and learn it. I'm right now taking a grade 12 math course. I'll be honest though it's been two years since I've taken a math course. I had some health issues and am now just finishing up highschool. Any recommendations for review? It's been so long I don't know where to start. I know how to do squares. It was a matter of figuring out what to do with the 55 infront of the = and whether I'm to multiply -5 by 55 first or square 55 first than multiply by -5. However I did some review on BEDMAS and I now know what to do first but I still don't know what to do with that 55 infront of the =
• May 7th 2011, 05:44 PM
Quacky
You have an incorrect statement. You are right to say that
$P(55)=-5(55)^2 + 550(55) - 5000$ and that this will give you your maximum.

This does not mean that $55=-5(55)^2 + 550(55) - 5000$. Think of $P(x)$ as another notation for ' $y$'. Obviously, you could use some review here to help you remember things like terminology and the order of operations - if you are struggling to recall grade 10 and 11 material, then how do you expect to cope with grade 12? (Wink)
• May 7th 2011, 05:51 PM
Arise01
So I don't need to move the 55 to the other side. It just acts as a notation. Ok so...

P(55) = -5(55)^2 + 550(55) - 5000
P(55) = -5(3025) + 30250 - 5000
P(55) = -15125 + 30250 - 5000
P(55) = 15125 - 5000
P(55) = 10125

I think I've done that right.
• May 7th 2011, 05:53 PM
Quacky
There we go.(Wink)
• May 7th 2011, 05:53 PM
mr fantastic
Quote:

Originally Posted by Arise01
55 = -5(55)^2 + 550(55) - 5000

I'm not sure how to calculate it out. Where do I start?

For this sort of arithmetic, clearly a student is expected to use a calculator ....
• May 7th 2011, 05:54 PM
Arise01
Quote:

Originally Posted by Quacky
Obviously, you could use some review here to help you remember things like terminology and the order of operations - if you are struggling to recall grade 10 and 11 material, then how do you expect to cope with grade 12? (Wink)

Got any tips besides going back and taking grade 10 and 11 math again? Or any things I could research?
• May 7th 2011, 05:58 PM
Arise01
Quote:

Originally Posted by mr fantastic
For this sort of arithmetic, clearly a student is expected to use a calculator ....

A calculator!... really... man I should have thought of that ... (sarcasm) lol
• May 7th 2011, 05:58 PM
mr fantastic
Quote:

Originally Posted by Arise01
Got any tips besides going back and taking grade 10 and 11 math again? Or any things I could research?

Such things are the responsibility of your teacher. That is who you should be speaking to.