Optimizing Product with Three Positive Numbers

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In summary, to find three positive numbers x, y, and z whose sum is 100 such that (x^a)(y^b)(z^c) is a maximum, use the constraint x+y+z=100 and the maximization function g(x,y)=x^a*y^b*(100-x-y)^c. Taking the partial derivatives of g with respect to x and y and setting them to zero, you can solve for x and y in terms of a, b, and c. These values can then be plugged into the constraint to find z.
  • #1
fk378
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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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  • #2
You most certainly haven't done this correctly!

I advise you to use Lagranfe multipliers.
 
  • #3
We haven't learned that yet.
 
  • #4
All right, then!

Post the equations you got after taking the partial derivatives and setting them to zero.
 
  • #5
x=(100a-100ay)/(c+a)
y=100c/a
Plugging this into x+y+z=100, I get z=-100c/a
 
  • #6
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!
 
  • #7
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.
 
  • #8
No, you don't. For one thing, 100y will not appear anywhere.
 

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