Question 1130815
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You mean *[tex \LARGE A\ =\  \frac{1}{2}R^2\sin\theta] where *[tex \LARGE R] is the length of one of the equal sides of the isosceles triangle and *[tex \LARGE \theta] is the vertex angle.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\(7\)^2\sin\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{49}{2}\sin\theta]


Any single angle of any triangle must be in the range *[tex \LARGE 0^\circ\ <\ \theta\ <\ 180^\circ].  The maximum value for the sine function in this range is 1 when *[tex \LARGE \theta\ =\ 90^\circ].  Since *[tex \LARGE \frac{49}{2}] is independent of the measure of the vertex angle, the maximum area must occur where *[tex \LARGE \sin\theta] is maximum.


Or you can use the Calculus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(\theta)\ =\ \frac{1}{2}R^2\sin\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dA}{d\theta}\ =\ \frac{R^2}{2}\cos\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dA}{d\theta}\ =\ 0\ \Right\ \cos\theta\ =\ 0\ \Right\ \theta\ =\ 90^\circ]


Therefore *[tex \LARGE A(90^\circ)] is a local extremum of the function


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2A}{d\theta^2}\ =\ -\frac{R^2}{2}\sin\theta]


Which is a negative value when *[tex \LARGE \theta\ =\ 90^\circ], therefore *[tex \LARGE A(90^\circ)] is a local maximum of the area function.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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