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# If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum, when the angle between them is $60^{\circ}$.

$(a)\;minimum\qquad(b)\;maximum\qquad(c)\;equal\qquad(d)\;None\;of\;these$

Let the hypotenuse of the right triangle be x, and the height be y
Hence its base is $\sqrt{x^2-y^2}$ by applying phythagorous theorem.
Hence its are = $\large\frac{1}{2} \times base \times height$
Area = $\large\frac{1}{2}$$\times \sqrt{x^2-y^2} \times y But it is given x+y = p(say) Substituting this in the area we get Area = \large\frac{1}{2}$$ \times \sqrt{(p-y)^2} - y^2 \times y$
$\large\frac{1}{2}$$y \sqrt{p^2+y^2-2py-y^2} = \large\frac{1}{2}$$y \sqrt{p^2-2py}$
Squaring on both the sides we get
$( Area)^2 = \large\frac{1}{4}$$y^2(p^2-2py) i.e., A = \large\frac{1}{4}$$ y^2(p^2-2py)$
$= \large\frac{1}{4}$$p^2y^2 = \large\frac{1}{2}py^3 For maximum or miniumu area \large\frac{dy}{dA}$$= 0$
Here the area of the triangle is maximum when $x =\large\frac{2p}{3}\: and \: y = \large\frac{p}{3}$
$\cos \theta = \large\frac{y}{x}$
$= \large\frac{\Large\frac{p}{3}}{2\Large\frac{p}{3}}$
$\therefore \cos \theta = \large\frac{1}{2}$
$\Rightarrow \theta = \Large\frac{\pi}{3}$ or $60^{\circ}$
Hence the area is maximum if the angle between the hypotenuse and the side is $60^{\circ}$
edited Apr 14, 2014