# Maximize the product xyz

1. Jul 9, 2008

### fk378

1. The problem statement, all variables and given/known data
Find three positive numbers x, y, and z whose sum is 100 such that (x^a)(y^b)(z^c) is a maximum.

2. Relevant equations
constraint: x+y+z=100
maximize: (x^a)(y^b)(z^c)

3. The attempt at a solution
First I replaced the z in the maximization problem with 100-y-z. Then I took the partial derivatives of the maximization function with respect to x and to y. Solving these, I got x=100. This implies that y=-z. But the question asks for all positive numbers. I don't know what else to do...any tips?

2. Jul 9, 2008

### arildno

You most certainly haven't done this correctly!

I advise you to use Lagranfe multipliers.

3. Jul 9, 2008

### fk378

We haven't learned that yet.

4. Jul 9, 2008

### arildno

All right, then!

Post the equations you got after taking the partial derivatives and setting them to zero.

5. Jul 9, 2008

### fk378

x=(100a-100ay)/(c+a)
y=100c/a
Plugging this into x+y+z=100, I get z=-100c/a

6. Jul 10, 2008

### arildno

This is wrong!

You are to differentiate:
$$g(x,y)=x^{a}y^{b}(100-x-y)^{c}$$
The partial derivative of g with respect to x becomes, using the product&chain rules:
$$\frac{\partial{g}}{\partial{x}}=ax^{a-1}y^{b}(100-x-y)^{c}-cx^{a}y^{b}(100-x-y)^{c-1}$$

Now, if this is set to zero, it means:
$$ax^{a-1}y^{b}(100-x-y)^{c}=cx^{a}y^{b}(100-x-y)^{c-1}$$
Assuming that all factors are non-zero, we may divide through, say in this manner:
$$a(100-x-y)=cx$$
Make a similar manipulation of the equation you gain from dg/dy!

7. Jul 10, 2008

### fk378

If you simplify the last equality you made, you get the same answer as I did for x.

8. Jul 10, 2008

### arildno

No, you don't. For one thing, 100y will not appear anywhere.