How do i solve this quadratic problem?
Image of question:
For part b) specifically, im not sure how they rearranged to get r = 400000 -1000(p -20)^2
- AlanLv 71 month agoFavorite Answer
I find the way they begin as dumb,
Why write a t = M - 1000p
t= 10000 + (30-p)1000
t = 40000 -1000p
so M = 40000
r= p(40000 -1000p) = 40000p -1000p^2
complete the square
r = -1000p^2 + 40000p = -1000(p^2 -40p)
(p-20)^2 = p^2 -40p + 400
p^2 -40p = (p-20)^2 -400
so replace p^2 -40p with " (p-20)^2 -400 "
r = -1000( (p-20)^2 -400)
r = 400000 -1000(p-20)^2
so this is vertex form
so the max is
p = 20
r = 400000
- AmyLv 71 month ago
(a) is simple. Fill in t = 10 000 and p = 30, then solve for M.
(b) you need to complete the square.
This reduces the equation from containing two p's to one.
r = p(M - 1000p)
= Mp - 1000p^2
= -1000(p^2 - M/1000 p)
= -1000(p^2 - M/1000 p + (M/2000)^2 - (M/2000)^2)
= -1000(p - M/2000)^2 + M^2/4000
(c) The equation r = A - B(p - c)^2 has a maximum at p=c (or a minimum if B is negative).