Math problem!!?!?:O?

Question

I would be really thankful if anyone would slove this..problem: divide the number 48 into three parts such that it's product is maximum..this problem is based on maxima and minima for functions of two variables..(second semester engineering mathematics)

Answer 1

divide it equally.

a + b + c = 48

maximise abc.

M(a, b) = abc = ab(48-a-b) = 48ab - a²b - ab²

M is the quantity to maximise.

dM/da = 48b - 2ab - ab² = 0

dM/db = 48a - 2ab - ab² = 0

subtract these two equations to get:

48(a-b) = 0

a = b

Now use symmetry to deduce a = c, and conclude that abc is maximised when a = b = c, for the constraint a + b + c = constant.

Hope this helps.