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°.

Let the hypotenuse of the right triangle be x and the height be y. Hence its base is √x²−y² by applying Pythagoras theorem. Hence its area = 1/2 × base × height
Area = 1/2 × √x²−y² × y
But it is given x + y = p (say)
Substituting this in the area, we get
Area = 1/2 × √(p−y)²−y² × y = 1/2y√p²−2py
Squaring on both the sides, we get
(Area)² = 1/4y²(p²−2py)
i.e., A = 1/4y²(p²−2py) = 1/4p²y²−1/2py³
For maximum or minimum area, dA/dy = 0
Here the area of the triangle is maximum when x = 2p/3 and y = p/3
cosθ = y/x = (p/3)/(2p/3) = 1/2
∴ θ = π/3 or 60°
Hence, the area is maximum if the angle between the hypotenuse and the side is 60°.