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danbone87
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[SOLVED] Finding the Max Value of a Function on a closed interval
Hi, I'm reposting this because it's a subsection to a larger question I had and I figured more people might be able to help with a new topic name.
anyway i have the equation (eq1) C(t) = C + a(e^.5t) + b(e^.9t) t>0 [3,12]
a,b and C are constants.
I realized from long ago that the max and mins of a function occur when y'=0
So I took the derivative of the function which gave me
C(t) = Css + a(e^.5t) + b(e^.9t) t>0
c'(t)= .5a(e^.5t) + .9b(e^.9t) = 0
i took the ln of the whole thing
and that gave me
[ln(.5a)*.5t] + [ln(.9b)*.9t] = 0
my question now is can use the property that ln(xy) = ln(x)+ln(y) to combine the two ln's and thereby combining the coeffiecients of t such that it is .5t+.9t = .14t
and the whole thing would be .14t*ln(.45ab)= 0
... that seems to be wrong considering that the t's cancel out but I can't think of much else...
Homework Statement
Hi, I'm reposting this because it's a subsection to a larger question I had and I figured more people might be able to help with a new topic name.
anyway i have the equation (eq1) C(t) = C + a(e^.5t) + b(e^.9t) t>0 [3,12]
a,b and C are constants.
The Attempt at a Solution
I realized from long ago that the max and mins of a function occur when y'=0
So I took the derivative of the function which gave me
C(t) = Css + a(e^.5t) + b(e^.9t) t>0
c'(t)= .5a(e^.5t) + .9b(e^.9t) = 0
i took the ln of the whole thing
and that gave me
[ln(.5a)*.5t] + [ln(.9b)*.9t] = 0
my question now is can use the property that ln(xy) = ln(x)+ln(y) to combine the two ln's and thereby combining the coeffiecients of t such that it is .5t+.9t = .14t
and the whole thing would be .14t*ln(.45ab)= 0
... that seems to be wrong considering that the t's cancel out but I can't think of much else...