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Finding the Max Value of a Function on a closed interval

  • Thread starter danbone87
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[SOLVED] Finding the Max Value of a Function on a closed interval

1. 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.


3. 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...
 

Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
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1. 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.


3. The Attempt at a Solution

I realized from long ago that the max and mins of a function occur when y'=0
Not quite. If the max and/or min occur in the interior of an interval, they must occur where y'= 0. The max and/or min may occur at the endpoints of the interval.

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
No, no no! Ln(x+ y) is NOT ln(x)+ ln(y)!!
You could write it as .5ae.5t= -.9be.9t and take logs of both sides of that. Be careful about signs. One of a and b must be negative but you don't know which one. Either .5ae.5t= -.9be.9t or -.5ae.5t= .9be.9t gives you positive numbers on each side so you can use logarithms. Try both and see what happens.

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...
Yes, that is wrong!
 
Dick
Science Advisor
Homework Helper
26,258
618
The 'taking the ln of the whole thing' is a completely wrong move. E.g. ln(0)=-infinity and there is no rule that says ln(a+b)=ln(a)+ln(b). Rearrange the equation first. Start by writing .5*a*e^(.5*t)=-.9*b*e^(.9*t). Now move all of the e^ type things over to one side and everything else over to the other. Now 'take the ln of the whole thing'.
 
HallsofIvy
Science Advisor
Homework Helper
41,732
893
1. 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.


3. The Attempt at a Solution

I realized from long ago that the max and mins of a function occur when y'=0
Not quite. If the max and/or min occur in the interior of an interval, they must occur where y'= 0. The max and/or min may occur at the endpoints of the interval.

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
No, no no! Ln(x+ y) is NOT ln(x)+ ln(y)!!
I recommend factoring out e.5t: e.5t(.5a+ .9be.4t)= 0. Since e.5t is never 0, you must have .5a+ .9be.4t= 0. That should be easy to solve for t. Of course, one of a and b must be negative for this to be true.


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...
Yes, that is wrong!
 
28
0
Should've kept those calc books :grumpy: thanks for the help guys
 
28
0
Just to clarify, I said that the ln(x*y) = ln(x) +ln(y) not ln(x+y)

that is correct isn't it?
 
Dick
Science Advisor
Homework Helper
26,258
618
That's correct. That's what you SAID. If you look back at what you actually USED, it was ln(x+y)=ln(x)+ln(y) (???) when you 'took the ln of the whole thing'.
 
28
0
yeah, I figured that. It just comes into play right after when I seprate these t's out so i wanted to double check. Thanks again.
 

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