Two sides of a Triangel have lengths of A and B, and the angle between them is Theta what value of Theta will maximixe the triagle's area? HINT
A=(1/2)AB sin Theta
so where do i start?
Actually you do not need derivatives.
When is,
$\displaystyle A(\theta)=\frac{1}{2}xy \sin \theta$ maximized?
When the sin of the angle is maximized, when is that?
When $\displaystyle \sin \theta=1$ (that is largest value of sine function).
That happens when $\displaystyle \theta=\frac{\pi}{2}$
The maximum area will be where the derivative of the area function is 0. (Note: we also get minima from this so we need to prove that we actually have a maximum.)
A' = (1/2)AB cos(theta)
Setting this to zero we get
cos(theta) = 0
which happens at theta = pi/2, 3*pi/2 rad.
The 3*pi/2 rad angle is ridiculously large for a triangle, so use theta = pi/2 rad. Is this point a local maximum? You can either do the second derivative test (ie show that A''(pi/2) < 0 ) or simply look at the graph of A(theta) at pi/2. It turns out that this is a maximum for the function.
-Dan