[High School- Algebra 2] Using Quadratics to Max. Profit

[Yoshiisland]

I'm having a lot of problems just writing the equation, let alone finding the answer.

A community theater sells about 150 tickets to a play each week when it charges $20 per ticket. For each $1 decrease in price, about 10 more tickets are sold. The theater has a fixed expense of $1500 a week.

Write the quadratic function that represent theater's weekly profit.

How much the theater should charge to max weekly profit.

[u84]

Duh, they should charge $17 each for 170 tickets. I'm dumb.

Note, I don't know what they mean when they say they want it in quadratic function.

[Relambrien]

I've never actually taken "Algebra 2," but I think my pre-calc class did pretty much the same thing, so I'll take a stab at this.

The profit is going to be (# of tickets)(price per ticket). In this case, the number of tickets is determined by 150 + 10x, where x is the number of dollars less than $20 that the theater charges. Which means that price per ticket is (20 - x)

So plugging those in you get

(150 + 10x)(20 - x), which can be expanded into a quadratic equation, then graphed to find at what x-value the function maximizes (or if you know any calculus, take the derivative then find where it is equal to zero).

I'll put my own work in white below this in case you want to check/verify/whatever:


(150 + 10x)(20 - x)

3000 - 150x + 200x - 10x^2

-10x^2 + 50x + 3000

Note: the next two lines involve calculus
-10x + 50 = 0
x = 5

$15

(150 + 10(5)) * (20 - 5)
(150 + 50) * (15)
200 * 15
3000

[u84]

Originally Posted by Relambrien

I've never actually taken "Algebra 2," but I think my pre-calc class did pretty much the same thing, so I'll take a stab at this.

The profit is going to be (# of tickets)(price per ticket). In this case, the number of tickets is determined by 150 + 10x, where x is the number of dollars less than $20 that the theater charges. Which means that price per ticket is (20 - x)

So plugging those in you get

(150 + 10x)(20 - x), which can be expanded into a quadratic equation, then graphed to find at what x-value the function maximizes (or if you know any calculus, take the derivative then find where it is equal to zero).

I'll put my own work in white below this in case you want to check/verify/whatever:


(150 + 10x)(20 - x)

3000 - 150x + 200x - 10x^2

-10x^2 + 50x + 3000

Note: the next two lines involve calculus
-10x + 50 = 0
x = 5

$15

(150 + 10(5)) * (20 - 5)
(150 + 50) * (15)
200 * 15
3000

So, what is your final answer to it? You still have to answer the question.

[Oni-Paranoia]

Relambrian speaks like my teachers

In other words just take what he first said and plug in X with possible solutions.
E.g. X = 15

(150 + 10x)(20 - x) = (150 + 10*15)(20-15)

(300)(5) = 1500

Like Relambrian, i dont get the quadratic crap (havent got that far in Alg 2 yet)

Edit: by the equation above im gonna guess the theater should charge $15 a week to make $1500. (don trust me for final anwser, this is a learning process for me as well)

[Relambrien]

Originally Posted by u84

So, what is your final answer to it? You still have to answer the question.

Well I wasn't going to do -all- of the work, just the bit the OP was having trouble with.

[Yoshiisland]

Ah ok. So all I need to is foil, and find the vertex.

Which will give me the should decrease the price by $2.50.

And it goes (sold-price)(people+cost)?

[u84]

Originally Posted by Yoshiisland

Ah ok. So all I need to is foil, and find the vertex.

Which will give me the should decrease the price by $2.50.

And it goes (sold-price)(people+cost)?

Well, It says decrease by $1, so it would have to be either $17 or $18, but yeah, you are on the right track. I don't know these funky equations, but I do know that $17 and $18 give 180 and 190 tickets, but $19+ and $16- give you a profit of $1540 and below. This leads me to believe that $17 and $18 are the only possible solutions.

[Gilly G]

Relambrien was on the right track, but made an error for the derivative

-10x^2 + 50x + 3000

Derivative should thus be:
-20x + 50

So then:
-20x + 50 = 0
x = 2.5

(150 + 10x)(20 - x) <=> (150 + 10(2.5))(20 - 2.5)
= 175 * 17.5
= 3062.5

Since 20 - x is the price the theatre should charge, we get $17.5 for maximum profit.

[Yoshiisland]

That's what I actually got from Ram using the vertex formula.

[QED Stepfiles]

Let us call x the amount of dollars UNDER $20 the theater charges for a ticket.

Therefore, number of tickets sold is (150+10x)

Price of the ticket is (20-x)

Thus, total REVENUE is (150+10x)(20-x)

Total profit is this minus the weekly expense, so it is

(150+10x)(20-x)-1500 = 3000 + 50x -10x^2 - 1500 = -10x^2 + 50x + 1500

That's your quadratic equation. To find the maximum, set the derivative to zero:

-20x + 50 = 0

x = 5/2

Therefore, the theater should charge 20 - 5/2 = 17.5 dollars per ticket to maximize profit. Of course, you should really check that this is a MAXIMUM of the function by taking the second derivative, plugging in 5/2, and making sure that the second derivative is negative...

The second derivative is

-20

which is negative always. So we're done. The answer is $17.50, and the quadratic function is

-10x^2 + 50x + 1500.

Note that the theater's fixed weekly response has NOTHING to do with the price they should charge. Their weekly fixed cost could have been 5 million dollars and the answer would still be $17.50. Mathematically this is because constants vanish when you differentiate them, but logically this makes sense since it is a FIXED cost and as such it must be paid no matter what, and thus should not have any bearing on how much they should charge for a ticket.

Hope this helps!

PS: Note that we had to make the (reasonable) assumption that the number of tickets sold as a function of price charged is approximately linear.

[NumiGamer]

Yoshiisland and Oni-Paranoia:
Apparently you're in Algebra 2 like me. How the hell are you that far in the book by now? I only just had a test on adding/subtracting/multiplying matrices and scalar multiplication on Friday!

*EDIT* Not looking forward to quadratic functions.
Does NOT look fun.

[x After Dawn x]

Quadratic functions are easier than **** guys, I go to one of the hardest high schools in the world and even I found it easy.

Just to let you know, yes, the vertex's y-value should represent the max revenue, or you could take the x-value of the vertex and put it in for x in the formula and you should get the same thing.

Another thing you can do is foil out the formula and convert it into the standard formula. The q value in a(x-q)+p should represent the number of increases / decreases (x-value) and the p value should represent the max revenue.

Also, if you want to find a certain revenue of something, just graph the revenue you want in y2 of your calculator and look at where the quadratic intersects with the constant. If the vertex hits exactly at the constant, then there is only one x-value (increase or decrease) to satisfy the formula. If it hits it twice, there are two x-values (but make sure that you're setting your window to + y values and + x values only because technically, if it went into the negatives, that would mean that you'd have to increase the price, and you should only look at this if your teacher says you can for that question). If it doesn't hit it at all, it means that no matter how many increases / decreases there are, it will never reach that revenue.

Hope this helps.

[Relambrien]

Originally Posted by Gilly G

Relambrien was on the right track, but made an error for the derivative

****, you're right. Mistype screwed me over for the rest of the problem.

I hate computers sometimes.

[sumzup]

As QED Stepfiles said, it doesn't matter what the costs are in order to maximize profit, but it's still important to note that Profit = Revenue - Costs, which implies the profit equation is what QED Stepfiles derived, not what Relambrien came up with.

[Oni-Paranoia]

Originally Posted by NumiGamer

Yoshiisland and Oni-Paranoia:
Apparently you're in Algebra 2 like me. How the hell are you that far in the book by now? I only just had a test on adding/subtracting/multiplying matrices and scalar multiplication on Friday!

*EDIT* Not looking forward to quadratic functions.
Does NOT look fun.

Im not that far yet, and i did matrices in Alg 1
Different states = Different curriculums
I just got done working with Functions x.x