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

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