Optimization word problem

In designing airplanes, an important feature is the so-called drag feature, that is, the retarding force exerted on the plane by air. One model measures drag by a function of the form F(v)=Av^2+(B/v^2)
where v is the velocity of the plane ad A and B are constants. Find the velocity (in terms of A and B) that minimizes F(v). Show that you have found the minimum rather than the maximum.


Please show steps! Thanks!

Min / Max occur at critical points, where the derivative of your function is 0.

f(v) = av^2 + b/v^2
f'(v) = 2av - 2b(v^(-3))

Setting the derivative to 0:
0 = 2av - 2b(v^(-3))
av = bv^(-3)
v^4 = b/a

This provides two real solutions:
v = (b/a)^(1/4)
v = -(b/a)^(1/4)

And we only care about the positive velocity.

Now to prove that this is a minimum not a maximum, we consider the second derivative of the drag function. If the graph is concave up at the critical point, then you have a local minimum.
f''(v) = 2a + 6b(v^(-4))

Since f''((b/a)^(1/4)) > 0, the graph is concave up, and v = (b/a)^(1/4) must be a local minimum.

I'm going to assume both A,B are positive. The problem seems to be false otherwise.

Compute the derivative.
F '(v) = 2Av - ( 2B / v^3 )

Set this equal to zero and solve for v.
v = (B/A)^(1/4)

Finally, compute the second derivative.
F '(v) = 2A + (6B / v^4)
This is always positive, so the v we found above is the minimum.