Let a and b be positive integers such that a+b = 20. What is the maximum value that (a^2)b can take?

We need to maximise the value of (a^2)b given that a+b = 20.The first step is to note that we can eliminate b from these equations to give a problem in just one variable. We have b = 20 - a, so (a^2)b = (a^2)(20 - a) = 20a^2 - a^3. To maximise 20a^2 - a^3, we can differentiate this expression and set it equal to 0 to find the turning point. This gives 40a - 3a^2 = 0(40 - 3a)a = 0 and as a = 0 gives a^2b = 0, this isn't the maximum so 40 - 3a = 0 and a = 13+1/3. However the question said that a and b must be integers and the closest integer values of a to this are 13 and 14 so the maximum is either a = 13 and b = 7 or a = 14 and b = 6. Computing (a^2)b in both of these cases gets 1183 and 1176. So the maximum value is 1183.