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Math Help - Very strange critical number problem...

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    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.
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    Quote Originally Posted by Kaitosan View Post
    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 ?
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    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.
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    Quote Originally Posted by Kaitosan View Post
    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.
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    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.
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    Quote Originally Posted by Kaitosan View Post
    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 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 f(a)<f(b)
    So for any x\geq 0, f(x)\geq f(0), meaning that f(0) is the minimum of f.


    Now, you can see, here, that 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.
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    Thanks bro.
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