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(B) Let the two equal sides of the triangle be $x$ and the angle between them be $2	heta$. From the geometry of the triangle,the height is $h = x \si

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$\frac{1}{2}x^2$

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(B) Let the two equal sides of the triangle be $x$ and the angle between them be $2	heta$. From the geometry of the triangle,the height is $h = x \sin 	heta$ and the base is $b = 2x \cos 	heta$. The area of the triangle is given by $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. $A = \frac{1}{2} 	imes (2x \cos 	heta) 	imes (x \sin 	heta) = x^2 \sin 	heta \cos 	heta$. Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we get $A = \frac{1}{2} x^2 \sin 2	heta$. To maximize the area $A$,we need to maximize $\sin 2	heta$. The maximum value of $\sin 2	heta$ is $1$,which occurs when $2	heta = 90^\circ$ or $	heta = 45^\circ$. Therefore,the maximum area is $A_{\max} = \frac{1}{2} x^2 (1) = \frac{1}{2} x^2$.

$A$ triangular garden is fenced on two sides,and the third side is a straight river bank. The two fenced sides have an equal length of $x$. What is the maximum area that can be enclosed by the garden?

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