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

Explanation

`b+a=k(constant) `
or `b+c=k`

Let’s take the side to be `BC = a`

`⇒a + b = k`
where k is a constant.

Using Pythagoras theorem, we get `c = sqrt(b^2-a^2)`
We want to maximize the area of the triangle.

Area `=1/2 base xx altitude`
`=1/2 a xx c`
we found that,`c=b^2-a^2`
Substituting it in the expression for area, we get
Area `=1/2 a xx sqrt(b^2-a^2)`
let A be the area,
so, `A^2=1/2 a^2xx(b^2-a^2)`
Also, from `a+b = k`, we get `b = k-a`
Substituting it in the expression for area, we get
`A^2=1/2 xx(k^2-2ka)xxa^2`
`=>A^2=14(k^2a^2-2ka^3)`
This is a function which gives the square of area.
Since k is a constant, we can say this is a function of a.
When the area is maximum, area square will also be maximum.
we will differentiate with respect to a and equate to zero.
`2A dA / da = 1 /4 (2ak^2-6ka^2)`

`dA /da = 0`
=> `2ak^2-6ka^2=0`

Or `a = k / 3`,
we neglect other values because side has to be positive

From `a+b = k`,
we get `b = k – k /3 = 2k / 3`

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