If the sum of the lengths of the hypotenuse and another side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between these sides is `pi/3dot`

Explanation

Let ABC be a right angle traingle in which side BC=x(say)
and hypotenuse AC=y(say).Given x+y=k (constant)
or , y=k-x
Now the area of triangle ABC is given by
`A=(1)/(2)BC AB =(1)/(2)xsqrt(y^(2)-x^(2))=(1)/(2)xsqrt[((k-x)^(2)-x^(2))]`
Let `u=A^(2)=(1)/(4)x^(2)(k^(2)-2kxc)`
`therefore du//dx=(1)/(2)k(kx-3x^(2))` and `d^(2)u//dx^(2)=(1)/(2)k(k-6x)`
For maxium or minimum or u,
`du//dx=0`
or x=`k//3`
When `x=k//3, d^(2)u//dx^(2)=(1)/(2)k(k-6xx(1)/(3)k)=-(1)/(2)k^(2)(-ve)`
Therefore, u,i.e A is maximum when `x=k//3` and when
`y=k-x=2k//3`
Now `costheta=BC//AC=x//y=1//2or theta=pi//3`
Hence the required angle is `pi//3`