Given: a right-angled triangle, such that sum of the lengths of its hypotenuse and side is given
To show: the area of the triangle is maximum when the angle between them is π/3
Explanation:
Let ΔABC be the right-angled triangle,
Let hypotenuse, AC = y,
side, BC = x, AB = h
Now sum of the side and hypotenuse is given,
⇒ x+y = k, where k is any constant value
⇒ y = k-x………..(i)
Now let A be the area of the triangle, then we know
Now substituting equation (i) in above equation, we get
Now substituting the value from equation (iii) into equation (ii), we get
Now applying the product rule, we get
Applying the power rule of differentiation on second part in above equation, we get
Now critical point is found by equating the first derivative to 0, i.e.,
⇒k2-2kx = kx
⇒k2 = 2kx+kx
⇒ k2 = 3kx
⇒ k = 3x
⇒ x = k/3
Again, differentiating equation (iv) with respect to x, we get
Applying the product rule of differentiation, we get
Now applying the power rule of differentiation, we get
Substituting x = k/3, in above equation, we get
Now from figure,
cos θ = x/y
Substituting the value of y = k-x from equation (i), we get
Hence the area of the triangle is maximum only when the angle between them is π/3.