This calculus problem is driving me crazy!?

Question

When the admission price for a baseball game was $6.00 per ticket, 36,000 tickets were sold. When the price was raised to $7.00, only 33,000 tickets were sold. Assume that the demand function is linear and that the variable and fixed costs for the ballpark owners are $0.20 and $85,000 respectively.

I'm supposed to find the profit P as a function of x, the number of tickets sold.

My profit equation is coming out with insane decimals. This is worth 8 points on my test and the test is in a few hours and I haven't slept yet due to THIS problem! PLEASE HELP! What do you get for your profit equation?

My teacher's question is what price would you set your ticket prices at to obtain maximum profit?

When find the slope of the 2 points (36000,6) and (33000, 7) I get:
-1/3000 and my end point slope formula looks like this:
Y = -1/3000 - 4

AHH I GOT IT! Thank you all very much! (I didn't just copy your answers I did it without looking after seeing where I got stuck.) You guys don't know how much you saved me! Now at least I get 4 hours of sleep before the test.

Answer 1

First, let's find the equation for the line through (36000, 6) and (33000, 7), because this relates the number of tickets sold to the price per ticket. The slope is (7-6)/(33000-36000)=-1/3000. We have
y=-1/3000 x + b, and we want to know b; let's plug in a point.
6 = (-1/3000)(36000) + b
6 = -12 + b
b=18

So, x tickets are sold at $y each if
y = 18-x/3000.

The total revenue from these ticket sales is the number of tickets times the price per ticket:

R=xy=x(18-x/3000)=18x-x^2/3000

Costs to the owners are 85000+.20x (fixed costs are just the number; variable costs are a cost per ticket--say, a ticket printing fee, or maybe a seat warming fee, or whatever).

Profit is revenue minus cost, or 18x-x^2/3000-85000-.20x
P = 17.8x-x^2/3000-85000

To maximize profit, we set its derivative to zero:

P' = 17.8 - x/1500 = 0
17.8 = x/1500
x = 17.8*1500 = 26700

This answers the question, but to go a little further: maximum profit is 17.8(26700)-(26700)^2/3000-85000
= 475260 - 237630 - 85000
= 152630

And the ticket price we use to achieve maximum profit is 18-26700/3000 = 18-8.9 = $9.10

Answer 2

OMG i have not finished this for a at the same time as, enable's attempt! f''(x) = 24x^2+2x+10 int(f''(x)) = 8x^3 + x^2 + 10x +C = f'(x) 8 *a million + a million + 10 +C = -6 ===> C = -25 re-write f'(x) = 8x^3 + x^2 + 10x -25 int (f'(x)) = 2 x^4 + a million/3 x^3 + 5 x^2 -25x +C = f(x) 2 + a million/3 + 5 -25 +C = -a million ===> C = 16.6667 f(x) = 2 x^4 + a million/3 x^3 + 5 x^2 -25x + 16.6667 i imagine we've close solutions u have divided C via 2 someplace not a biggy or am i incorrect?

Answer 3

Well dude the first equation you should have gotten for profit verses number of tickets is.
P=-3000x+54000

I'm guessing your teacher has you integrate it..
the integral of that would be:

Something = -1500x^2+54000x+constant

I don't know the steps that were taught in your class for this type of word problem. So if you want to repost the equations you've calculated and then people can check your math.

Answer 4

As straightforward as possible:

(T)ickets = 54000-3000(P)rice
(R)evenue = P x T = 54000P-3000P^2

(C)osts = 85000 + 0.2T = 85000 + 10800 - 600P = 95800 - 600P

Profit = R - C = -3000P^2 + 54600P - 95800

Differentiate this and set to 0 to maximize:

-6000P + 54600 = 0 ==> P = 9.10

Answer 5

D(p) = -3000(p)+54000 (point slope formula)

P(x) = D(p) (p) -.2(D(p))-85000

thus p(x) = (p-.2)(-3000(p) + 54000) -85000

simplify this, then perform a D2 test to find the maximum.