The function is obviously defined only for `v gt 0` and is continuously differentiable on this interval. When `v` approaches zero the function tends to `+oo,` when v tends to `+oo,` the function also tends to `+oo.`
Thus it must have at least one local (and global) minimum and it is reached at the point(s) where `P'(x) = 0.` Let's solve this equation:
`P'(x) = -17 v^-2 + 3*10^-3*v^2 = 0.`
This is equivalent to `17 v^-2 = 3*10^-3*v^2,` or `v^4 = 17/3*10^3.`
The only solution is `v = root(4)(17/3*10^3) approx` 8.7 (m/s). This is the answer.
Take the derivative of `P(v)` .
`P'(v)=-17v^-2+3*10^-3v^2`
Set `P'(v)` equal to zero and solve for the critical values.
`-17v^-2+3*10^-3v^2=0`
`-17+3*10^-3v^4=0`
`3*10^-3v^4=17`
`v^4=17/3*10^3`
`v^4=17/3*10^3`
`v=(17/3*10^3)^(1/4)~~8.676 m/s`
We can check graphically that this is indeed a minimum.
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