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 b/w them is pie/3.
consider the following triangle ABC
Let AC= y and BC= x and x + y = k ( k is a constant)
Let A be the area of triangle then
...... (1)
After differentiating (1) w.r.t x we get
...... (2)
putting (2) equal to 0 we get x= k/3
again differentiating (2) we get
...... (3)
put dA/dx=0 and x=k/3 in (3)
Thus, A is maximum when x=k/3
Now, x = k/3
⇒y=k-k/3=2k/3
therefore cosθ=x/y
⇒ cosθ=(k/3)/2k/3=1/2
⇒θ = π/3