Working with Lagrange multipliers, reducing gradients is okay, right?

103 Views Asked by At

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.

1

There are 1 best solutions below

0
On BEST ANSWER

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 $\lambda$.