SOLUTION: the population with a certain disease for the time 0 < t < 20 is determined by the function p(t) = 1/100 (20t^3 - t^4). When is the population a maximum? When is the disease will c

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Question 1194793: the population with a certain disease for the time 0 < t < 20 is determined by the function p(t) = 1/100 (20t^3 - t^4). When is the population a maximum? When is the disease will cease?
Found 2 solutions by Alan3354, MathLover1:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
the population with a certain disease for the time 0 < t < 20 is determined by the function p(t) = 1/100 (20t^3 - t^4). When is the population a maximum? When is the disease will cease?
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p(t) = 1/100 (20t^3 - t^4)
p'(t) = (1/100)*(60t^2 - 4t^3) 1st derivative
(1/100)*(60t^2 - 4t^3) = 0
60t^2 - 4t^3 = 0
t^2*(60 - 4t) = 0
t = 15 is the max
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p(t) = 1/100 (20t^3 - t^4) = 0
(20t^3 - t^4) = 0
t^3*(20-t) = 0
p(t) = 0 at t=0 and at t=20

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

the population with a certain disease for the time 0+%3C+t+%3C+20 is determined by the function
p%28t%29+=+%281%2F100%29+%2820t%5E3+-+t%5E4%29
Critical points ,max and min, are points where the function is defined and its derivative is zero or+undefined.

When is the population a maximum?
max:

f'%28t%29=%281%2F100%29+%2820t%5E3+-+t%5E4%29+=+-%281%2F25%29%28t-15%29%2At%5E2
-%281%2F25%29%28t+-15%29%2At%5E2=0
at t%5E2=0=>t+=+0+
and
t+-+15=0 =>t=15
p%280%29+=+%281%2F100%29+%2820%2A0%5E3+-+0%5E4%29=0
p%2815%29+=+%281%2F100%29+%2820%2A15%5E3+-+15%5E4%29=675%2F4

then
Global Maximum is at (15, 675%2F4 )
Global Minimum is at (0,+0+)

When is the disease will cease?
p%28t%29+=0
%281%2F100%29+%2820t%5E3+-+t%5E4%29=0
20t%5E3+-+t%5E4=0
20t%5E3+=+t%5E4
20+=+t%5E4%2Ft%5E3
t=20
the disease will cease at t=20