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If the sum of the lengths of the hypotenuse 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 π/3.

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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.

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