If the sum of lengths of hypotenuse and a side of a right angled triangle is given, show that area of triangle is maximum, when the angle between them is `pi/3`.

Explanation

Let, In `Delta ABC, angleB=90^(@)`
Let AB = x and AC = y
Then x+y = k
Which is given a constant.
`BC= sqrt(AC^(2)-AB^(2))`
`=sqrt(y^(2)-x^(2))`
`= sqrt((k-x)^(2)-x^(2))`
`=sqrt(k^(2)-2kx)`
`" Now area of "Delta ABC`
`Delta = 1/2 xx AB xx BC`
`=1/2 xx x xx sqrt(k^(2) - 2 kx)`
`rArr Delta^(2) = 1/4 x^(2) (k^(2)-2 kx)`
Let` Delta^(2) = S`,
`:. S=1/4 k (kx^(2)-2x^(3))`
`rArr (dS)/(dx)= 1/4 k(2 kx - 6x^(2))`
For maxima/minima
` (dS)/(dx) = 0`
`rArr 2kx - 6x^(2) = 0`
`rArr x= k/3`
and `(d^(2)S)/(dx^(2)) = 1/4 k(2k-12x)`
` "at "x = k/3`
`(d^(2)S)/(dx^(2)) = 1/4 k (2k-4k) =-(k^(2))/(2) lt 0`
`:." At " x = k/3,*`S is maximum
`rArr" At " x = k/3, Delta^(2)` is maximum
`rArr" At "x= k/3, Delta` is maximum
and ` y=k-k/3=(2k)/3`
If `angle BAC = theta, ` then
`cos theta = (AB)/(AC) = x/y = (k//3)/(2k//3) = 1/2`
`rArr theta = pi//3` ,brgt Therefore the required angle is `pi//3` for maximum area.