Consider the following for the two (02) items that follow: Let ABC be a triangle right-angled at B and AB + AC = 3 units. What is the maximum area of the triangle?

Step-by-step solution

Let AB = c , BC = a , and AC = b . Given that the triangle is right-angled at B , we have a^2 + c^2 = b^2 . We are given AB + AC = 3 , which means c + b = 3 b = 3 - c . Substituting b into the Pythagorean theorem: a^2 + c^2 = (3 - c)^2 a^2 + c^2 = 9 - 6c + c^2 a^2 = 9 - 6c a = 9 - 6c The area of the triangle is = 1 2 ac = 1 2 c 9 - 6c . To maximize the area, we can maximize its square, S = ^2 : S = 1 4 c^2(9 - 6c) = 9 4 c^2 - 3 2 c^3 Differentiating S with respect to c and equating to zero: dS dc = 9 2 c - 9 2 c^2