Quadratic function for determining profit.

[This name sucks]

p(x)=-30x^2+550x-400

p(x) represents profit and x represents ticket costs.

What is the maximum possible profit and how many sales are required to break even.

I haven't done this shit in like 5 years and I need to quickly relearn it and google isn't helping me that much.

I'm assuming I need to find the vertex for the max profit and the zeros for the break even but im not quite sure and want a few more opinions.

[Sunshine]

Are you sure the initial equation is correct?

And I'd suggest trying what you said, then see if it looks reasonable.

[Collegeguy]

ah, applied calculus. Been so long since I did that, and I remember it being a boringly easy concept.

[Catta]

I get a maximum of 2121 in profits as a max with the tickets costing 9,167.

And yes that was just finding the maximum on the graph it creates.


How much you need to sell to break even is weird imo. You have to sell tickets for 400 but as the price is changing...
If talking about breaking even with the newly found maximum profit price of tickets it would be 17,57.
Solve(0=-30x^2+550x-400,x=9,167) = x = 17,57

[Tepesch]

p(x)=-30x^2+550x-400

So you need the maximum p(x) depending on x. Derive p(x):

the general rule is:

d/dx x^n = n * x^(n-1)

so

d/dx p(x) = - 2 * 30 * x + 550

the maximum is at

d/dx p(x) = 0

-60*x + 550 = 0 -> x = 9.16

and therefore

p(x=9.16) = 2120

[Deleted]

p(t)=-30x^2+550x-400

p(t) represents profit and x represents ticket costs.

What is the maximum possible profit and how many sales are required to break even.

I haven't done this shit in like 5 years and I need to quickly relearn it and google isn't helping me that much.

I'm assuming I need to find the vertex for the max profit and the zeros for the break even but im not quite sure and want a few more opinions.

If you don't know what you're talking about don't reply at all please.

1. Use derivative to find the maximum points in the equation
p'(t) = -60t + 550
p'(t0) = -60t + 550 = 0
t = 9.17

2. Break-even is when profit is 0.
p(t)=-30t^2+550t-400 = 0
t = 17.57 or t = 0.76

[procne]

1. Use deriving to find the maximum points in the equation
p'(t) = -60x + 550
p'(t0) = -60x + 550 = 0
x = 9.17

Maybe I'm nitpicking, but if p(t) = -30x^2+550x-400
then it's a constant. Function defines t as variable, not x
Therefore p'(t) = 0

[Deleted]

Maybe I'm nitpicking, but if p(t) = -30x^2+550x-400
then it's a constant. Function defines t as variable, not x
Therefore p'(t) = 0

Yes you're nitpicking. I just blindly copied from OP.

[Auloria]

In case you think in logic and not math:

Think about plotting that line on a graph. What does the maximum point look like? It looks like the graph going up then turning around and going back down. That's why you want the derivative, which is just the slope of the line. The derivative is positive when the graph is going up, then gets smaller and passes through zero at the maximum and turns negative when the graph goes back down. So you take the derivative, you set it equal to zero, and you find the value(s) of x that make it work. However, you want to make sure that you are looking at a maximum, not a minimum, where the derivative is also zero. To do that you look at the second derivative, which tells you that the slope is increasing or decreasing.

You break even when p(x) = 0, so you just solve the original equation for the zeros.

When you are solving for the zeros of an equation, you are trying to turn Ax^n + Bx^n-1 + Cx^N-2 + ... into something that looks like (x-A)(x-B)(x-C).. = 0. Then the solution is that if any of the pieces are 0, you get p(x) = 0. So, if x = A, or if x = B, etc.

[This name sucks]

I get a maximum of 2121 in profits as a max with the tickets costing 9,167.

And yes that was just finding the maximum on the graph it creates.


How much you need to sell to break even is weird imo. You have to sell tickets for 400 but as the price is changing...
If talking about breaking even with the newly found maximum profit price of tickets it would be 17,57.
Solve(0=-30x^2+550x-400,x=9,167) = x = 17,57

How did you come to the price of the ticket for the maximum possible profit.

Just solve for x?

Everything else you said matches what I ended up working out.

[Deleted]

How did you come to the price of the ticket for the maximum possible profit.

Just solve for x?

Everything else you said matches what I ended up working out.

Cattaclysmic has incomplete answers. There are two breakeven points.

He got the maximum profit by solving the derivative of the profit function and then solving x = 0.

[This name sucks]

Cattaclysmic has incomplete answers. There are two breakeven points.

He got the maximum profit by solving the derivative of the profit function and then solving x = 0.

17.57 and .73 are both break even points, I knew that.


Thanks.

[Blueobelisk]

...got some wanna be smarties in here. Damn calc 1 level heroes before I showed up.

But yeah they're right.

[Catta]

Cattaclysmic has incomplete answers. There are two breakeven points.

He got the maximum profit by solving the derivative of the profit function and then solving x = 0.

No. I got em by solving p(t) = 0

[Deleted]

I'm with Cattaclysmic on this one.
I don't know of what use may the second break-even point be.

[Catta]

How did you come to the price of the ticket for the maximum possible profit.

Just solve for x?

Everything else you said matches what I ended up working out.

I dont have my calculator in my house right now so i cant check it till tomorrow but looking at the equation in my post i can see something is wrong as it sets an x value and then gives a new one. So ive written something wrong there.

[This name sucks]

I dont have my calculator in my house right now so i cant check it till tomorrow but looking at the equation in my post i can see something is wrong as it sets an x value and then gives a new one. So ive written something wrong there.

....shit...

[Deleted]

I dont have my calculator in my house right now so i cant check it till tomorrow but looking at the equation in my post i can see something is wrong as it sets an x value and then gives a new one. So ive written something wrong there.

Generally speaking, you maximize profits with respect to x (with x being the quantity), since revenues and costs are both functions of x-considering prices are fixed. You can then let maximum profits be a function of price solely!

....shit...

You've just fucked up your variables from what it seems:P

Btw, what the hell does "t" stand for?

[Catta]

....shit...

Eh, i think its just because i wrote ,x=9,167 instead of ,x
Maybe thats why i didnt get two break even points

---------- Post added 2012-12-05 at 12:51 AM ----------

Generally speaking, you maximize profits with respect to x (with x being the quantity), since revenues and costs are both functions of x-considering prices are fixed. You can then let maximum profits be a function of price solely!


You've just fucked up your variables from what it seems:P

But here x is ticket costs and not quantity.

[Deleted]

But here x is ticket costs and not quantity.

Well, if that's so, he's not gonna find any ticket quantity, just the price at which his profits are maximized.