# Thread: Very strange critical number problem...

1. ## Very strange critical number problem...

Find T that causes the minimum value of P(t) when P'(t) = 1 - 3e^(0.2sqrt(T))

Then find the minimum value of P(t)

If you look carefully, it seems like it's impossible to exactly express T, but only sqrt(T)...... ? So I'm confused about that.

2. Originally Posted by Kaitosan
Find T that causes the minimum value of P(t) when P'(t) = 1 - 3e^(0.2sqrt(T))

If you look carefully, it seems like it's impossible to exactly express T, but only sqrt(T)...... ?
If you can express sqrt(T), then you can express T, by squaring it

Also, are t and T the same ?

3. Haha. That's why I said "look carefully."

Input the value of sqrt(T) and T in the rate equation and you will see that the results are completely different.

4. Originally Posted by Kaitosan
Haha. That's why I said "look carefully."

Input the value of sqrt(T) and T in the rate equation and you will see that the results are completely different.
I don't get it at all -_-

What are sqrt(T) and T ?

Is it P(t)=1-3e^(0.2sqrt(T))
or P(T)=1-3e^(0.2sqrt(T)) ?

In the first case, it's a nonsense, since it's not even a function of t.

5. t = T I apologize for failing to be consistent.

You will see that, after solving for p'(t)'s zero, that sqrt(t) = negative value. That is the true zero of p'(t) but if you square it to get an expression for t, it doesn't work as a zero for p'(t) at all.

6. Originally Posted by Kaitosan
t = T I apologize for failing to be consistent.

You will see that, after solving for p'(t)'s zero, that sqrt(t) = negative value. That is the true zero of p'(t) but if you square it to get an expression for t, it doesn't work as a zero for p'(t) at all.
Okay, now I understand better ^^

If it's never equal to 0, it doesn't mean it has no minimum value

Think about it : suppose f is defined for $\displaystyle x\geq 0$. If the derivative is strictly positive, then the function is strictly increasing. This would mean that the minimum value of f is attained for the smallest possible value of x, that is 0.
Because if f is increasing, if a<b, then $\displaystyle f(a)<f(b)$
So for any $\displaystyle x\geq 0$, $\displaystyle f(x)\geq f(0)$, meaning that $\displaystyle f(0)$ is the minimum of f.

Now, you can see, here, that $\displaystyle 1-3e^{0.2 \sqrt{T}}<0$ (by taking the second derivative for example)
So the function is strictly decreasing.
So the minimum value of f is attained at the maximum value of T.
If it's infinity, there is no minimum.

7. Thanks bro.