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?

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- Oct 25th 2006, 05:27 PMcyberdx16App of derivatives... Max and min problem solving
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? - Oct 25th 2006, 05:33 PMThePerfectHacker
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 $\displaystyle \theta=\frac{\pi}{2}$ - Oct 25th 2006, 05:33 PMtopsquark
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