Let ABC be a right angle triangle in which side BC = x(say) And hypotenuse AC = y(say) Given x + y = k(const) = y = k - x Now the area of the triangle ABC is given by A =
1
2
BC . AB =
1
2
x√(y2−x2) =
1
2
x√[(k−x)2−x2] For max. or min. of u ,
du
dx
= 0 = x =
k
3
(∵ x ∊ 0) When x =
k
3
,
d2u
dx2
=
1
2
k (k - 2k) = −
1
2
k2 < 0 u , i.e. A , is max. when x =
k
3
and when y = k - x = 2k/3 Now cosθ = BC/AC = x/y = 1/2 = θ = π/3 Hence the required angle is π / 3