# Maximum/Minimum Problem

1. Jun 22, 2005

### Samael

I'm having some problems with this following min/max problem.

$$C= 0.25t{e^-^0^.^3^t}$$

I am being asked to find the Maximum and when it occurs (t)

Using the PR I obtained:

$$0.25t . {-0.3{e^-^0^.^3^t} + {e^-^0^.^3^t} . 0.25$$

Later by factoring

$$0.25 . e^-^0^.^3^t[(t - 0.3)]$$

Now because SP/TP at dy/dx = 0

I find one to give 0.3, but am not sure what to do with the other one $$0.25 . e^-^0^.^3^t$$

When I make it equal to zero I get an answer which is wrong, I'm guessing the error is either in my factorising or transposing the above problem?

2. Jun 22, 2005

### Zurtex

Think about the graph e-t, does it ever actually equal 0?

3. Jun 22, 2005

### steven187

hello there

C=o.25te^(-0.3t)
dC/dt=0.25e^(-0.3t)-0.075te^(-0.3t)=0
take one term to the right hand side then log both sides then use some logarithms rules and you should eventually get t by it self and t=0.2/0.075

take care

steven

4. Jun 22, 2005

### dextercioby

Your notation is rather awkward.Use brackets or the commands

"\cdot" or "\times" which produce

$$\cdot$$ , $$\times$$

,but don't put the multiplicative dot like that "."

Daniel.

5. Jun 22, 2005

### Zurtex

Although actually, as it is taught in England, . is the multiplication dot where as $\cdot$ is the decimal place. Akward

6. Jun 22, 2005

### dextercioby

You drive with the wheel on your right and you used Henry the VIII-th small finger to define the inch,so why am i not surprised...?

Daniel.

7. Jun 22, 2005

### Samael

As it is in my textbook "." is what I am taught.

Also those values you got don't seem to be the maximum (or minimum) values, nor can I obtain them by substituting them into $$f(x)$$.

Edit: My mistake, my factorising was done wrong. Should've looked like this:

$$f'(x) = e^-^0^.^3^t(-0.075t + 0.25)$$

Which gives the max as :

3.33 when letting the 2nd pair = 0

Last edited: Jun 22, 2005
8. Jun 22, 2005

### wisredz

So what's wrong. You solved the problem. You can never find a minimum on the interval (-inf,inf). It is unbounded below. Anyway Rolle's Theorem states that if a local or global maximum/minimum of a function f(x) at a point c, at that point the slope of the curve is zero. That is to sayf'(c)=0. You found the formula for the derivative. Suppose that there's a maximum at a point c. Using Rolle's theorem we get f'(c)=0. Now if you equal the formukla to zero and make some calculations, c turns out to be 3,333333... Put that number c into the formula of f(x)(you seek the extreme values of this function) and you have the absolute maximum of the function which is 0,306566. Any problems?

I wrote the whole solution because you already have done the most imprtant part.

9. Jun 22, 2005

### wisredz

:) I think I was writing when you edited your last post. You still don not have the maximum of the function. You just found the point where the extrema occurs... Plug that value into the first equation...

10. Jun 22, 2005

### Samael

Sorry, I was being asked to find when it occured hence we found the X point to be 3.333333 and then substituted it back into $$f(x)$$ to find the corresponding y value.

My problem was in how to use the PR correctely, but that's been figured out now. Thanks again. :)