The lengths of the sides of a triangle are 30-2x , 30-3x , and 5x (where x > 0 ). Let M be the maximum possible area of this triangle, and let this maximum occur at x = . An infinite geometric progression has its first term as and its sum to infinity as M 15 2 . The value of 20r , where r is the common ratio of the geometric progression, is equal to

Step-by-step solution

Let the sides of the triangle be a = 30-2x , b = 30-3x , and c = 5x . The semi-perimeter s is: s = (30-2x) + (30-3x) + 5x 2 = 30 Using Heron's formula, the area is: = s(s-a)(s-b)(s-c) = 30(30 - (30-2x))(30 - (30-3x))(30 - 5x) = 30(2x)(3x)(30-5x) = 180x^2(30-5x) = 30 6x^2 - x^3 To maximize the area, we maximize the function inside the square root, f(x) = 6x^2 - x^3 . Differentiating with respect to x and setting it to zero: f'(x) = 12x - 3x^2 = 0 3x(4 - x) = 0 Since x > 0 , we have x = 4 . Thus, = 4 . The maximum ar