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I am employing the method of Lagrange multipliers to determine a maximum.

As part of this, I arrive at the following equation involving two gradients and the parameter $\lambda$, as is common for this method:

$$\left< 3z, 6, 3x \right> = \lambda \left< 2x, 4y, 2z\right>.$$

Now, I need to proceed with the calculation. In order to simplify the arithmetic I want to simplify this equation. Specifically, I wonder if it an admissible simplification to first 'reduce' the gradients?

That is, can I consider the equation $$ \left< z, 2, x \right> = \lambda \left< x, 2y, z \right>$$ instead of the original one?

As the gradient is just a vector leading to the 'steepest ascent,' it makes sense that the direction is what matters... right?

But I am not certain if it is legal to first 'reduce' the gradients, and would appreciate a clarification on this point.

quid
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armor
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    I find this to be an (extremely) odd close, personally - the question is quite understandable. And yes, since you're just solving the system $$\begin{align}3z &= \lambda 2x \ 6 &= \lambda 4y \ 3x &= \lambda 2z\end{align},$$ and multiplying/dividing/ both sides of an equation yields an equivalent equation, you're good to go if you're just trying to solve a system of equations. – pjs36 Jul 18 '16 at 19:02
  • @pjs36 the question is slightly different, I believe, as OP does not multiply by the same constant on both sides, but different ones. So the two systems of equations are not completely equivalent; there is a rescaling of the $\lambda$. – quid Jul 18 '16 at 22:34

1 Answers1

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Say you have:

$$\left< 3z, 6, 3x \right> = \lambda \left< 2x, 4y, 2z\right>$$

Clearly this is equal to:

$$\left< z, 2, x \right> = \frac{2\lambda}{3} \left< x, 2y, z\right>$$

You can let $\lambda^* = 2\lambda/3$ and proceed as usual. As you said, the direction is the most important part of Lagrange Multipliers, for the most part $\lambda$ is simply a scaling factor. Keep in mind that the physical meaning of $\lambda^*$ will be different than $\lambda$.

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George V. Williams
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