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

2 Answers

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  • Alan
    Lv 7
    1 month ago
    Favorite 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 

    (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).

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