Can someone please help me with this calculus problem?

Question

An object with weight W is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle A with the plane, then the magnitude of the force is:
F= uW / ( u sinA + cosA)
where u is a positive constant and where 0 <=A<=pi/2.
Show that F is minimized when tanA =u.

Thank you!

Answer 1

First of all, you have to realise that

F= uW / ( u*sinA + cosA)

will be minimised by maximising (u*sinA + cosA) since it is the denominator (and a bigger denominator gives you a smaller number).

So let G(A) = u*sinA + cosA, be the function you're trying to maximise.

The minimums/maximums will be found when G'(A) = 0

G'(A) = u*cosA - sinA

Using the fact that tanA = sinA/cosA, if tanA = u, then sinA = ucosA

Plugging this result in G'(A), we get, G'(A) = u*cosA - u*cosA = 0

The last thing left to show is that this is a maximum and not a minimum (it could be either) for G(A). For that, you need to study the second derivative and relalise that it will be a maximum if the second derivative is negative around the desired point:

G''(A) = -u*sinA - cosA

Using the fact that tanA = sinA/cosA, if tanA = u, then sinA = ucosA

Plugging this result in G''(A), we get, G''(A) = -(u^2)*cosA - cosA which is obviously negative if u is positive.

Thus, you have proven that if tanA = u, F will be minimised.

Answer 2

F´= uW * 1/(u sin A +cos A )^2 * (-u cos A + sin A)
F´= 0
-u cos A =-sin A so tan A = u
The sign of Fín the interval
cos A( tan A -u) is ------u++++++ as tanA goes from 0 to infinity

Answer 3

Question is wrong.Please recheck.