Question

using f(p)= 180-0.3p^2 where p is the price in dollars and f(p) is the number of items sold. at what price will maximum revenue be generated? rounded to the nearest cent

Answer 1

that's weird..i got a maximum turning point at p=0 and f(p)=180

Answer 2

I think i did it wrong, ill check

Answer 3

okay thank you

Answer 4

I got p=0 again, what's the answer though?

Answer 5

$0 just looks wrong...so i dont know

Answer 6

If something is free 180 items will be sold. Lol.

Answer 7

I just did f'(p) = -0.6p
When f'(p)=0, p=0 and f(p)=180
And then i found the nature of the turning point... and it's a max

but you're trying to generate max revenue? why is everything gonna be free then

Answer 8

f(p) should be revenue. As we are trying to maximize this function.
p should be items sold.
I think....

Answer 9

yeah that's what puzzled me...ballards are you sure you typed out the Q correctly

Answer 10

here it is again straight from the test im trying to correct

Answer 11

The demand function for a certain comodity is given by f(p)= 180 - 0.3p^2, where p is the price on dollars and f(p) is the number of items sold.

#d) at what price will maximum revenue be generated? Round to the nearest cent

Answer 12

I think p is the price you pay to manufacture the item... so if you pay $0 then you still manage to sell 180 items and thus you generate max. revenue? Considering that you've made no loss? I don't know, ha.

Answer 13

ha ha ha thanks for trying though

Answer 14

Do you have answers for it the sheet you're doing?

Answer 15

for the*

Answer 16

no it was a test we have to correct for partial credit

Answer 17

this is what i did and she only counted off 3 instead of 5

Answer 18

162 = 180-.3p^2
-180 -180
-18= -.3p^2
/-.3 /-.3

60=p^2
p=7.75