Finding the Max Value of a Function on a closed interval

[danbone87]

[SOLVED] Finding the Max Value of a Function on a closed interval

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...

 

[HallsofIvy]

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

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]

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]

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

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!

 

[danbone87]

Should've kept those calc books :grumpy: thanks for the help guys

 

[danbone87]

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]

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'.

 

[danbone87]

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.