Thread: max/min on an interval #2

A swimming pool is treated periodically to control the growth of bacteria. Suppoe that t days after a treatment, the concentration of bacteria per cubic cm is C(t)=30t^9-240t+500. Find the lowest concentration of bacteria during the first week after the treatment.

Ok.. what I did.. I think i did right, but the back of the book.. just wondering if they are wrong. Just making sure, please confirm with me.

My work is below:
c'(t)=60t-240

critical numbers
c'(t)=0 when 60t-240=0
t=4

domain
0<=t<=7

test
f(4)=0
f(0)=-240
f(7)=180

therefore the lowest concentration of bacteria during the first week after the treatment is 0 cm^3 on the 4th day

That is my answer. However, the back of the book says.. 20

Originally Posted by skeske1234

A swimming pool is treated periodically to control the growth of bacteria. Suppoe that t days after a treatment, the concentration of bacteria per cubic cm is C(t)=30t^9-240t+500. Find the lowest concentration of bacteria during the first week after the treatment.

C(t) = 30t^2 - 240t + 500 instead of 30t^9 ... ?

Ok.. what I did.. I think i did right, but the back of the book.. just wondering if they are wrong. Just making sure, please confirm with me.

My work is below:
c'(t)=60t-240

critical numbers
c'(t)=0 when 60t-240=0
t=4

domain
0<=t<=7

test
f(4)=0 C(4) = 20
f(0)=-240 C(0) = 500
f(7)=180 C(7) = 290

therefore the lowest concentration of bacteria during the first week after the treatment is 0 cm^3 on the 4th day

That is my answer. However, the back of the book says.. 20

correct ... C(4) = 20 bacteria/cm^3

Hello, skeske1234!

A swimming pool is treated periodically to control the growth of bacteria.
Suppoe that days after a treatment,
the concentration of bacteria per cm³ is: .
Find the lowest concentration of bacteria during the first week after the treatment.

My work: .

Domain: .

This is correct . . . the rest is wrong.

You were testing values into the derivative, not the C-function.

. .

Edit: Once again, skeeter is too fast for me . . .
.

When you've got t = 4 you need to substitute 4 for t in the original equation, not its derivative, to get the concentration at that point.