Let's start by solving this problem when there are only two positive numbers involved, and then see whether we can apply the same technique when there are three positive integers.
Let the two positive integers be x and y. Then x + y = 100, and xy = the product.
Let's eliminate x. Solve x + y = 100 for x: x = 100 - y. Now subst. this last result into P = xy: P = product = (100 - y)(y) = 100y - y^2
Differentiating, dP/dy = 100 - 2y. Set this = to 0 and solve for y: -2y = -100, and y = 50. Since x + y = 100, x is thus also = to 50.
Solution set: (50,50).
Now suppose that three positive integers add up to 100, and that we want to maximize their product.
Then x + y + z = 100. Let's maximize f(x,y,z) = xyz (the product of x, y and z).
Since x + y + z = 100, we can eliminate z by solving x + y + z = 100 for z and subst. the result back into f(x,y,z) = xyz:
We get f(x,y) =xy(100-x-y), a function of two variables instead of three.
I won't go through the entire procedure of maximizing a function in three variables, but will get you started:
Find the 'partial of f with respect to x' and then the 'partial of f with respect to y'. Set each of these partial derivatives = to 0:
f = 0 = (partial of xy(100-x-y) with respect to x x
= xy(partial of 100-x-y with respect to x) + (100-x-y)(partial of xy with respect to x)
= xy(-1) + (100-x-y)(y)
We must set this partial = to 0: -xy+100y-xy-y^2 = 0
-2xy + 100y - y^2 = 0
or y(-2x + 100 - y) = 0
of which y=0 is one solution and in which -2x + 100 - y = 0
You must now go through the same procedure with respect to the partials with respect to y.
If you'd like to continue this discussion, please respond with questions and comments.
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