Given a + b = 20, find the maximum value of ba^2.

In the question we are given that a + b = 20 where a and b are positive integers. Remember this.First let y = a^2 b. This is a good starting point as we have an equality now which we can mess with.We want to find a maximum value of a function, so we want to determine the values of a and b such that the derivative, dy/da = 0 or dy/db = 0. Since we have a + b = 20, we say b = 20 - a and substitute this into our expression for y giving:y = a^2 b = a^2 (20 - a) = 20a^2 - a^3Differentiating with respect to a gives:dy/dx = 40a - 3a^2 = 0Solving for a gives: a = 40/3This value, however, is not an integer (which is required of a by the question). We solve this problem by considering the closest integers to 40/3 which are a=13 and a=14.For a=13, b = 20-a = 20-13 = 7 so y = a^2 b = (13)^2 (7) = 1183For a =14, b = 20-a = 20-14 = 6 so y = a^2 b = (14)^2 (6) = 1176.1183 is the higher value of the 2 so we can conclude that a = 13 and b = 7Done!