# Verify that the sum of three quantities x, y, z

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1. Jun 9, 2017

### Elias Waranoi

1. The problem statement, all variables and given/known data
Verify that the sum of three quantities x, y, z, whose product is a constant k, is maximum when these three quantities are equal.

2. Relevant equations
w = x + y + z
k = x * y * z

3. The attempt at a solution
Assuming that my understanding of the question is correct i.e. that we want the maximum of w and x*y*z has to be equal to a constant k then I'm pretty lost. If they mean the maximum as the largest value of w possible then it's definitely not when x = y = z.

If k = 1 then if x = y = z would give w = 3. But if k = 1 and x = 100, y = 0.1, z = 0.1 then k is still 1 and w = 100.2.

If they mean maximum as in where the rate of change of w is 0 then that doesn't make sense either because even though the rate of change of w is 0 when x = y = z, w is at it's lowest so that would make that point into a minimum.

The only thing I can think of is that they wrote maximum instead of minimum.

2. Jun 9, 2017

### WWGD

Haven't you seen the method of Lagrange multipliers?

3. Jun 9, 2017

### Elias Waranoi

Never heard of it. I'm working with the Calculus made easy book and I haven't encountered it there.

4. Jun 9, 2017

### WWGD

What kind of techniques have you seen/used to solve these types of problems?

5. Jun 9, 2017

### Elias Waranoi

This chapter was the first and only chapter on partial differentation. They showed that to find the maxima and minima of a multivariable function, the derivative of all partial derivatives have to be zero.

6. Jun 9, 2017

### ehild

You are right. The sum should be minimum.

7. Jun 9, 2017

### WWGD

Ah, sorry, I missed that, I thought they were asking for a minimum.

8. Jun 9, 2017

### Ray Vickson

The result is true for a minimum sum. There is no maximum sum under those conditions, because we can write the sum as
$$S = x + y + \frac{k}{xy},$$
and so can have $S \to +\infty$ by taking $x,y \to 0+$.

9. Jun 9, 2017

### Elias Waranoi

In that case, k = xyz, w = x + y + z = k/(yz) + k/(xz) + k/(xy)
∂w/∂x = -k/(x2z) - k/(x2y) = -y/x - z/x = 0 ∴ y = -z
∂w/∂y = -k/(y2z) - k/(y2x) = -x/y - z/y = 0 ∴ z = -x
∂w/∂z = -k/(z2y) - k/(z2x) = -x/z - y/z = 0 ∴ x = -y

Now this confuses me, because y = -z therefore x = -y = z. Now I have x = z and z = -x? Am I allowed to do this?

I tried testing with values and w doesn't even seem to be a minimum when x = y = z considering that if x = 1, y = -1, z = -1 then w = -1

10. Jun 9, 2017

### WWGD

What
constraints do you have on x,y,z ? Are they positive?

11. Jun 9, 2017

### Ray Vickson

Please use the appropriate reply button that identifies which post you are replying to. I cannot tell what you are commenting on in your response above.

For non-negative values of $x,y,z$ the point $(x,y,z) = (c,c,c)$ (with $c = k^{1/3}$ and $k > 0$) really is a provable minimum.
However, if you allow for some negative values of the variables, the sum has no finite minimum for the same reason I outlined in #8 above: you can have
$$S = x + y + \frac{k}{xy} \to -\infty$$
by having $x \to 0+$ and $y \to 0-$ (again, when $k > 0$).

Last edited: Jun 9, 2017