omg i cant get this damn math problem

[ManBearPig]

At a price of $8 per ticket, a theater group can fill every seat in their 1750 seat hall. For every additional dollar charged for admission, the number of tickets sold drops by 55

What ticket price maximizes revenue? Round your answer to the nearest cent.
price ?

i know that 1095 seats are sold at that price, i just have no idea how to find the amount they cost!

 

[ManBearPig]

 

[Inspector Jihad]

graph it and find where the two intersect.


edit: is the 1095 seats given in the problem?

 

[marketsons1985]

What kind of math class is this for?

 

[ManBearPig]

Jihad, i had to solve for the 1095 seats, and that was correct. market, its for calc.

thanks guys

 

[Inspector Jihad]

ok, i think i got it.

(1750 - 55x) = 1095

x = 11.9

ticket price = 11.9 + 8

price = $19.90

 

[Jhill]

Do they have to pay for parking?

 

[ColdFusion718]

Originally posted by: Jhill
Do they have to pay for parking?

 

[ManBearPig]

jihad thats not right, i got that too and i tried it

 

[marketsons1985]

The way to do it without knowing 1095 beforehand is this:

You know that the revenue equals cost * number of tickets sold. The but the number of tickets sold equals 1750 - 55*(cost-8). Thus revenue = cost * (1750-55*(cost-8)) for prices above 8. Find the derivative d/dcost of the revenue equation, and solve for that = 0. That is:

solve d/dc(c*(1750-55*(c-8)))=0 for c.

 

[Inspector Jihad]

Originally posted by: marketsons1985
The way to do it without knowing 1095 beforehand is this:

You know that the revenue equals cost * number of tickets sold. The but the number of tickets sold equals 1750 - 55*(cost-8). Thus revenue = cost * (1750-55*(cost-8)) for prices above 8. Find the derivative d/dcost of the revenue equation, and solve for that = 0. That is:

solve d/dc(c*(1750-55*(c-8)))=0 for c.

u be readin dem books n shit

 

[marketsons1985]

Originally posted by: Inspector Jihad

Originally posted by: marketsons1985
The way to do it without knowing 1095 beforehand is this:

You know that the revenue equals cost * number of tickets sold. The but the number of tickets sold equals 1750 - 55*(cost-8). Thus revenue = cost * (1750-55*(cost-8)) for prices above 8. Find the derivative d/dcost of the revenue equation, and solve for that = 0. That is:

solve d/dc(c*(1750-55*(c-8)))=0 for c.

u be readin dem books n shit

<------Math major that just got done tutoring someone on a similar problem...

 

[ManBearPig]

im too stupid to work out the equation you gave me lol.

cost is 8+x, right? so itd be...nevermind lol

 

[Leros]

This is calculus. You create an equation for the cost. Differentiate it and solve for where the derivate is 0. At every point where the derivate is zero, you have either a max or a min. You can test the two spots very close to both sides to see which it is. Obviously you are looking for a max.

diagram

 

[DrPizza]

What course are you taking?

A non-calculus approach to these optimization problems is to graph sets of inequalities. The optimal solution will be at one of the intersections.

 

[TecHNooB]

Originally posted by: marketsons1985
The way to do it without knowing 1095 beforehand is this:

You know that the revenue equals cost * number of tickets sold. The but the number of tickets sold equals 1750 - 55*(cost-8). Thus revenue = cost * (1750-55*(cost-8)) for prices above 8. Find the derivative d/dcost of the revenue equation, and solve for that = 0. That is:

solve d/dc(c*(1750-55*(c-8)))=0 for c.

Psh, way to make it unnecessarily hard to read!

It's basically:

Total Revenue = Ticket Price * (1750 Seats - 55 ( Ticket Price - 8))

or

Y = X * (1750 - 55 (X - 8))
to find maximum Total Revenue or Y you take the derivative of that equation and find the critical points (hence the d/dc). The answer turns out to be 19.90 which you stated earlier to be wrong.

 

[imported_quasi]

Ah, optimization with calculus...I have a test over this tomorrow.

 

[rocadelpunk]

Originally posted by: marketsons1985

Originally posted by: Inspector Jihad

Originally posted by: marketsons1985
The way to do it without knowing 1095 beforehand is this:

You know that the revenue equals cost * number of tickets sold. The but the number of tickets sold equals 1750 - 55*(cost-8). Thus revenue = cost * (1750-55*(cost-8)) for prices above 8. Find the derivative d/dcost of the revenue equation, and solve for that = 0. That is:

solve d/dc(c*(1750-55*(c-8)))=0 for c.

u be readin dem books n shit

<------Math major that just got done tutoring someone on a similar problem...

If you wanted to do it algebraically

We know that Rev = #tix sold * cost of a ticket

The cost of a ticket is a linear function (we can tell b/c it's slope is -55...i.e. rise/run (rise one dollar you run negative 55 tickets )
Graphing cost of ticket on x-axis v.s. #tickets sold on y axis gives y = -55x + 2190 (given two pts: (8,1750), (9, 1695) and then just finding equation of a line)

so rev = x* (-55x + 2190)

which will give you a parabala and since the coefficient in front of the x^2 is negative we know its vertex is a max, so it'd just be -b/2a i.e. -2190/2*-55 = 19.909

so we know to maximize profit we need to sell each ticket for x= $19.91

To find the number of seats sold you'd just plug it into your y = -55x + 2190

 

[chuckywang]

Originally posted by: Kazaam
At a price of $8 per ticket, a theater group can fill every seat in their 1750 seat hall. For every additional dollar charged for admission, the number of tickets sold drops by 55

What ticket price maximizes revenue? Round your answer to the nearest cent.
price ?

i know that 1095 seats are sold at that price, i just have no idea how to find the amount they cost!

Let x = amount over $8 that theatre charges.

Find the max of (8 + x)(1750 - 55x)

 

[chuckywang]

Originally posted by: marketsons1985

Originally posted by: Inspector Jihad

Originally posted by: marketsons1985
The way to do it without knowing 1095 beforehand is this:

You know that the revenue equals cost * number of tickets sold. The but the number of tickets sold equals 1750 - 55*(cost-8). Thus revenue = cost * (1750-55*(cost-8)) for prices above 8. Find the derivative d/dcost of the revenue equation, and solve for that = 0. That is:

solve d/dc(c*(1750-55*(c-8)))=0 for c.

u be readin dem books n shit

<------Math major that just got done tutoring someone on a similar problem...

A small change of variables, and this problem can be done in your head.

 

[Tiamat]

max and min are the derivative tests.

 

[Parasitic]

Why is there an increasing number of threads on people asking the ATOT population to do homework for them?

 

[TheoPetro]

Originally posted by: Parasitic
Why is there an increasing number of threads on people asking the ATOT population to do homework for them?

because calc is the hard for a lot of people. Try sitting in a stats class that forces you to use tables instead of calc because 98% of the people in there have trouble with college algebra and intuitive calc is like rocket surgery (or brain science) to them. Its WAY more frustrating than "hey help me with homework." I just wish I remembered more of my calc and ODE.

 

[MegaVovaN]

Originally posted by: TheoPetro

Originally posted by: Parasitic
Why is there an increasing number of threads on people asking the ATOT population to do homework for them?

because calc is the hard for a lot of people. Try sitting in a stats class that forces you to use tables instead of calc because 98% of the people in there have trouble with college algebra and intuitive calc is like rocket surgery (or brain science) to them. Its WAY more frustrating than "hey help me with homework." I just wish I remembered more of my calc and ODE.

Rocket science and brain surgery you mean?
And whats up with college algebra? My college alg class in community college is easy, I and several people from same class have A+ averages! Although I have seen some 50s and 60s in the pack of scantrons

 

[dighn]

Originally posted by: Parasitic
Why is there an increasing number of threads on people asking the ATOT population to do homework for them?

- it's a couple of months into school semester
- we solve their problems for them