Lagrange Multipliers

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This article discusses Lagrange multipliers, a topic of multivariable calculus.


Useful in optimization, Lagrange multipliers, based on a calculus approach, can be used to find local minimums and maximums of a function given a constraint. Suppose there is a continuous function $f(x,y)$ and there exists a continuous constraint function on the values of the function $c = g(x,y)$. The method of Lagrange multipliers states that, to find the minimum or maximum satisfying both requirements ($\lambda$ is a constant):

\[\dfrac{\partial f}{\partial x} = \lambda \dfrac{\partial g}{\partial x}\] \[\dfrac{\partial f}{\partial y} = \lambda \dfrac{\partial g}{\partial y}\]

The method can be extended to multiple variables, as well as multiple constraints. If we had a continuous function $f(x, y, z)$ and the constraint $c = g(x, y, z)$, we would just have another equation relating the partial derivatives with respect to $z$ through a factor of $\lambda$. Similarly, if we have another constraint $d = h(x, y, z)$, then we would add the partial with respect to each variable to their respective equation with another factor $\mu$. Thus, we have the general form for Lagrange multipliers for the function $f(x_1, x_2,...,x_n)$ bounded by a certain number of constraints $c_1 = g_1(x_1, x_2,..., x_n)$, $c_2 = g_2(x_1, x_2,..., x_n)$,...,$c_m = g_m(x_1, x_2,..., x_n)$:

\[\nabla f = \lambda_1 \nabla g_1 + \lambda_2 \nabla g_2 + ... + \lambda_m \nabla g_m\]

Where $\nabla$ is the del operator, which when applied to a scalar function $f$ results in a vector with each component the partial in that component's direction, representing total change.

It is important to remember that sometimes there maybe a function that is bounded on some interval. In such cases, Lagrange multipliers may give a result, but the answer may not be the one this method results. After using this method, it is imperative to check the endpoints and also plug back in the numbers resulted from this method to verify the final answer. Also, sometimes the function may be over-constrained and there maybe no point that satisfies all requirements.


This method can become extremely difficult to use when there are several variables and constraints, but it is very effective and useful in some situations, including some contest math problems. Consider the following question:

A rectangular prism lies on the $xy$ plane with one vertex at the origin. The vertex that does not share a face with this vertex at the origin lies on the plane bounded by the points $(0, 0, 4)$, $(0, 8, 0)$, $(16, 0, 0)$. Find the maximum volume of the box.

Solution: The volume of the box is given by the equation $V(x, y, z) = xyz$. Because the way the box is described, the point that lies on the plane given determines the volume of the box. Thus we need only need to consider that point. The constraint we are given for that point is given by the plane $x + 2y + 4z = 16$, since that is the plane that goes through the three points mentioned. Let us call that function $16 = g(x, y, z)$. Now we can begin to find our partial derivatives:

\[\dfrac{\partial V}{\partial x} = yz\] \[\dfrac{\partial V}{\partial y} = xz\] \[\dfrac{\partial V}{\partial z} = xy\] \[\dfrac{\partial g}{\partial x} = 1\] \[\dfrac{\partial g}{\partial y} = 2\] \[\dfrac{\partial g}{\partial z} = 4\]

We equate the corresponding partials through a constant factor, and we also use our original constraint: \[x + 2y + 4z = 16\] \[yz = \left(\lambda \right)1\] \[xz = \left(\lambda \right)2\] \[xy = \left(\lambda \right)4\]

Multiplying each side of the last three equations by the variable not in the equation, we have: \[xyz = \left(\lambda \right)1x\] \[xyz = \left(\lambda \right)2y\] \[xyz = \left(\lambda \right)4z\]

We can equate these to result: $x = 2y = 4z$

Plugging back into the constraint, we have: \[x + x + x = 16\] \[x = \dfrac{16}{3}\] \[y = \dfrac{8}{3}\] \[z = \dfrac{4}{3}\]

Thus, the maximum volume, given by $xyz$, is: $\boxed{\dfrac{512}{9}}$



Find the maximum area of a rectangular prism with a surface area of $624$.


Prove that \[0\le yz+zx+xy-2xyz\le \frac{7}{27}\], where $x,y$ and $z$ are non-negative real numbers satisfying \[x+y+z=1\].

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