# still cant get the answer....

• Mar 15th 2010, 02:07 PM
mike21
Hi I cant solve these two optimization problems...

1. suppose that 20,000 fans will go to a ball game when the price of a ticket is \$5, and that 500 fewer fans will go each \$1 increase in price. What should the ticket price be in order to maximize revenue?

2. compute the max profit when the demand function is p(x)=x^2-3x+2 and the total cost function is c(x)=2x^3/3-1/2x^2-2x. Recall that r(x) = xp. Enter just a reduced fraction of form a/b.

Can anyone solve these?

Thanks
- for the first one I set up in y-y1 = m(x-x1) form. with the points being (5,2000) and (6,19500). It came out to be y=-500x+22500. I then did R(x)=xp R(x)=x(-500x+22500) and distributed the x. Then i took the first derivitve and solved for 0. I got 22.5 but I think I solved for something else because that price is too high.
- As far as the second one I tried just plugging the two formulas in the formula p(x) = R(x)-c(x), taking the first derivitive and then solving for 0, but Im lost on it and really dont know what to do.
• Mar 15th 2010, 02:31 PM
apcalculus
Quote:

Originally Posted by mike21
Hi I cant solve these two optimization problems...

1. suppose that 20,000 fans will go to a ball game when the price of a ticket is \$5, and that 500 fewer fans will go each \$1 increase in price. What should the ticket price be in order to maximize revenue?

2. compute the max profit when the demand function is p(x)=x^2-3x+2 and the total cost function is c(x)=2x^3/3-1/2x^2-2x. Recall that r(x) = xp. Enter just a reduced fraction of form a/b.

Can anyone solve these?

Thanks
- for the first one I set up in y-y1 = m(x-x1) form. with the points being (5,2000) and (6,19500). It came out to be y=-500x+22500. I then did R(x)=xp R(x)=x(-500x+22500) and distributed the x. Then i took the first derivitve and solved for 0. I got 22.5 but I think I solved for something else because that price is too high.
- As far as the second one I tried just plugging the two formulas in the formula p(x) = R(x)-c(x), taking the first derivitive and then solving for 0, but Im lost on it and really dont know what to do.

2. compute the max profit when the demand function is p(x)=x^2-3x+2 and the total cost function is c(x)=2x^3/3-1/2x^2-2x. Recall that r(x) = xp. Enter just a reduced fraction of form a/b.

Revenue'(x) = 3x^2-6x+2
Cost'(x) = 2x^2 - x - 2

Set them equal, then simplify:

x^2 - 5x + 4 = 0

(x-1)(x-4) = 0

Two production levels are candidates for maximum profit. Pick the one where the second derivative of profit is negative (concave down, local maximum, by the second derivative test).

Good luck!

P'(x)