Homework Help: Pre-Calculus

Posted by Stacy on Saturday, November 19, 2011 at 4:18pm.

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)^2-120(3)+800
500-600+800 = 700

20(8)^2-120(8)+800
1280-960+800 = 1120

20(0)^2-120(0)+800 = 800

20(9)^2-120(9)+800 =1340

minimum = 700

but this is incorrect. What am I doing wrong and how do I solve it?

Pre-Calculus - Steve, Saturday, November 19, 2011 at 6:20pm
First off, t^2 - 6t + 40 does not factor into (t-5)(t-8) 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
Pre-Calculus - Stacy, Saturday, November 19, 2011 at 7:17pm
Thank you Steve. But i'm still a little confused. The answer in my answer list shows minimum as 620 at t=3.?
Pre-Calculus - Steve, Sunday, November 20, 2011 at 6:30am
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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