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fk378
Jul9-08, 05:57 PM
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?

arildno
Jul9-08, 06:02 PM
You most certainly haven't done this correctly!

I advise you to use Lagranfe multipliers.

fk378
Jul9-08, 06:26 PM
We haven't learned that yet.

arildno
Jul9-08, 06:34 PM
All right, then!

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

fk378
Jul9-08, 08:12 PM
x=(100a-100ay)/(c+a)
y=100c/a
Plugging this into x+y+z=100, I get z=-100c/a

arildno
Jul10-08, 03:15 AM
x=(100a-100ay)/(c+a)
y=100c/a
Plugging this into x+y+z=100, I get z=-100c/a

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!

fk378
Jul10-08, 09:30 AM
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!

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

arildno
Jul10-08, 09:33 AM
No, you don't. For one thing, 100y will not appear anywhere.