PreCalculus
posted by Stacy .
This is application question.
Bacterial Control. If t days after treatment the bacteria count per cubic centimeter in a body of water is given by C(t) = 10t^2  120x + 800
0 <= t <= 9
Thn in how many days will the jcount be a minimum?
I got this result:
20(t^2  6 + 40)
20(t  5)(t  8)
t  5 = 0 t  8 = 0
t = 5 t = 8
f(5) = 700
f(8) = 1120
f(0) = 800
f(9) = 1340
20(3)^2120(3)+800
500600+800 = 700
20(8)^2120(8)+800
1280960+800 = 1120
20(0)^2120(0)+800 = 800
20(9)^2120(9)+800 =1340
minimum = 700
but this is incorrect. What am I doing wrong and how do I solve it?

PreCalculus 
Steve
First off, t^2  6t + 40 does not factor into (t5)(t8) That would be t^2  13t + 40.
A parabola ax^2 + bx + c reaches its minimum where x = b/2a = 120/20 = 6
C(6) = 10*36  120*6 + 800 = 440 
PreCalculus 
Stacy
Thank you Steve. But i'm still a little confused. The answer in my answer list shows minimum as 620 at t=3.?

PreCalculus 
Steve
In that case, I'm also confused. If
C(t) = 10t^2  120x + 800
C(3) = 10*9  120*3 + 800
= 90  360 + 800
= 530
Is the function correct?
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