Optimizing Product with Three Positive Numbers

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Homework Help Overview

The problem involves finding three positive numbers x, y, and z that sum to 100, while maximizing the expression (x^a)(y^b)(z^c). The context is rooted in optimization within the framework of algebra and calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to substitute z in the maximization problem and takes partial derivatives, leading to a contradiction regarding the positivity of the variables. Some participants question the correctness of the original poster's approach and suggest using Lagrange multipliers, while others request the equations derived from the partial derivatives.

Discussion Status

The discussion is ongoing, with participants providing feedback on the attempted solutions and suggesting further exploration of the derivatives. There is a focus on clarifying the steps taken and the implications of the derived equations, but no consensus has been reached regarding the correct approach.

Contextual Notes

Some participants note that Lagrange multipliers have not been covered in the original poster's coursework, which may limit the methods available for solving the problem. There is also a concern about the validity of negative values for z derived from the equations presented.

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Homework Statement


Find three positive numbers x, y, and z whose sum is 100 such that (x^a)(y^b)(z^c) is a maximum.


Homework Equations


constraint: x+y+z=100
maximize: (x^a)(y^b)(z^c)


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?
 
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You most certainly haven't done this correctly!

I advise you to use Lagranfe multipliers.
 
We haven't learned that yet.
 
All right, then!

Post the equations you got after taking the partial derivatives and setting them to zero.
 
x=(100a-100ay)/(c+a)
y=100c/a
Plugging this into x+y+z=100, I get z=-100c/a
 
fk378 said:
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:
[tex]g(x,y)=x^{a}y^{b}(100-x-y)^{c}[/tex]
The partial derivative of g with respect to x becomes, using the product&chain rules:
[tex]\frac{\partial{g}}{\partial{x}}=ax^{a-1}y^{b}(100-x-y)^{c}-cx^{a}y^{b}(100-x-y)^{c-1}[/tex]

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

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

Now, if this is set to zero, it means:
[tex]ax^{a-1}y^{b}(100-x-y)^{c}=cx^{a}y^{b}(100-x-y)^{c-1}[/tex]
Assuming that all factors are non-zero, we may divide through, say in this manner:
[tex]a(100-x-y)=cx[/tex]
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.
 
No, you don't. For one thing, 100y will not appear anywhere.
 

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