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.
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.
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.