# how to setup the Lagrange multiplier for contraints such as $0\leq x \leq 1$

I need to find minimum of objective function $Q=f(x,y,z)$ inside a cube region. Hence the constraints are in the form $0 \leq x \leq 1$ and $0 \leq y \leq 1$ and $0 \leq z \leq 1$. Should the Lagrangian be

$$L = Q + \lambda_1(x) + \lambda_2(1-x) + \lambda_3(y) + \lambda_4(1-y) + \lambda_5(z) + \lambda_6(1-z)$$

All the examples I've seen have constraints on one side, such as $x\leq 1$, but what to do when the constraints have inequality on both sides? In general, how do I constraint the $Q$ for minimization inside 3D object such a cube.

An example $Q= n-z-x-y+x^2+y^2$ where $n$ is a number. and I want to find its min in a cube, hence constraints are the sizes of each of the three edges as given above. Just need to know how to write the constraints.

That's $1 + 6 + 12 + 8 = 27$ different cases. In general, you would have to solve each case, and reject all solutions that are not in the cube or on its surface. But perhaps your $Q$ allows some simplifications. Is $Q$ fixed? Or perhaps the second derivative matrix is positive definite?
• $Q$ is fixed. It is function of $x,y,z$, but not linear. it has terms in it such as $x^2$ and $y^2$. How can I then write the constraints such that they indicate Q is only inside the cube? Let us forget the surfaces and corners for now. Don't I need to use inequality on both sides to do this? strange that I am not able to find one example like this so far. – Steve H Dec 8 '14 at 11:31
• It would help if you told us what $Q$ is. In general, you can't use Lagrange multipliers to minimise over ranges like this; but perhaps the form of $Q$ admits simplifications. – TonyK Dec 8 '14 at 14:10
• fyi, just added an example of $Q$. – Steve H Dec 8 '14 at 19:31