Yemi asked in Science & MathematicsMathematics · 1 month ago

# How do i solve this quadratic problem?

Image of question:

https://imgur.com/a/md4BTv2

For part b) specifically, im not sure how they rearranged to get r = 400000 -1000(p -20)^2

Relevance
• Alan
Lv 7
1 month ago

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

(b)

r= p(40000 -1000p)  = 40000p  -1000p^2

complete the square

r =  -1000p^2  + 40000p  = -1000(p^2 -40p)

now

(p-20)^2  =  p^2 -40p  + 400

so

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

(c)

so this is vertex form

so the  max is

when

p = 20

r = 400000

• Amy
Lv 7
1 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).