Lagrange's Method of Multipliers
This method is used to find Maxima and Minima of a function in two or more than two variables called as Objective function. The variables are independent but are connected by a relation, called as Constraint function. Below are the steps to use this method:
Consider a function \(f(x,y,z) \) be an objective function of 3 variables \(x,y,z\) and the variables are connected by the constraint \(\phi(x,y,z) = 0 \)
Step 1 : Identify the objective function: \(f(x,y,z) \)
Step 2 : Identify the constraint function: \(\phi(x,y,z) = 0 \)
Step 3 : Find Lagrangian (Auxillary) function
\[ F(x,y,z) = f(x,y,z) + \lambda \phi (x,y,z) \longrightarrow (1) \]
\[ \lambda \longrightarrow \text{ non-zero } \longrightarrow \text{ Lagrange multiplier } \]
Step 4 : Partially differentiating equation (1) w.r.t. \(x,y,z\) and equate them to 0
\[ \frac{\partial F}{\partial x} = 0, \hspace{5pt} \frac{\partial F}{\partial y} = 0,\hspace{5pt} \frac{\partial F}{\partial z} = 0 \]
Step 5 : Solve the above 3 equations together with \(\phi(x,y,z) = 0 \) to find the stationary points
Step 6 : Find nature of the stationary points by using method discussed in Maxima and Minima i.e. \(rt - s^2 \). Here also lies its drawback: Lagrange's method does not give the nature of the stationary points.
Step 7 : Find maximum or minimum value of the function
A quick example will help you understand better.
Note : In our syllabus, we have to solve for only two variables.
Sample Problem 1
Find the minimum value of \(x^2 + y^2 + z^2 \) subject to the condition \(xyz = a^3 \).
Solution
Here, objective function is : \( f(x,y) = x^2 + y^2 + z^2 \longrightarrow (1) \)
Constraint function is : \(\phi(x,y) = xyz - a^3 \longrightarrow (2) \)
By Lagrange's method:
\[ F(x,y,z) = f(x,y,z) + \lambda \phi (x,y,z) \]
\[ = x^2 + y^2 + z^2 + \lambda (xyz - a^3) \longrightarrow (3) \]
Partially differentiating equation (3) w.r.t. \(x,y,z\)
\[ \frac{\partial F}{\partial x} = 2x + \lambda yz \longrightarrow (4) \]
\[ \frac{\partial F}{\partial y} = 2y + \lambda xz \longrightarrow (5) \]
\[ \frac{\partial F}{\partial z} = 2z + \lambda xy \longrightarrow (6) \]
Equating the equations to 0
\[ \frac{\partial F}{\partial x} = 2x + \lambda yz = 0 \longrightarrow (4) \]
\[ \frac{\partial F}{\partial y} = 2y + \lambda xz = 0 \longrightarrow (5) \]
\[ \frac{\partial F}{\partial z} = 2z + \lambda xy = 0 \longrightarrow (6) \]
Here you can solve for \(x,y,z\) and then for \(\lambda\), and get your stationary points, but there is also a short method for this:
Multiplying equation (4) by \(x\), (5) by \(y\) and (6) by \(z\)
\[ 2x^2 + \lambda xyz = 0 \longrightarrow (7) \]
\[ 2y^2 + \lambda xyz = 0 \longrightarrow (8) \]
\[ 2z^2 + \lambda xyz = 0 \longrightarrow (9) \]
Now on subtracting (8) from (7)
\[ 2x^2 + \lambda xyz - 2y^2 - \lambda xyz = 0 \]
\[ x = y \]
On subtracting (9) from (8)
\[ 2y^2 + \lambda xyz - 2z^2 - \lambda xyz = 0 \]
\[ y = z \]
\[ \Rightarrow x = y = z \]
Now using the constraint equation,
\[ x^3 = a^3 \Rightarrow x = a \]
Now as it is given in the question, we have to find minimum value, so we don't need to find nature of the stationary point.
So the minimum value of the function
\[ f(x,y) = a^2 + a^2 + a^2 = 3a^2 \]
Sample Problem 2
Find the point upon the plane \(ax + by + cz = p \) at which the function \[ f = x^2 + y^2 + z^2 \] has a minimum value and find the minimum value of \(f\).
Solution
Here, objective function is : \( f(x,y) = x^2 + y^2 + z^2 \longrightarrow (1) \)
Constraint function is : \(\phi(x,y) = ax + by + cz - p \longrightarrow (2) \)
By Lagrange's method:
\[ F(x,y,z) = f(x,y,z) + \lambda \phi (x,y,z) \]
\[ = x^2 + y^2 + z^2 + \lambda (ax + by + cz - p) \longrightarrow (3) \]
Partially differentiating equation (3) w.r.t. \(x,y,z\)
\[ \frac{\partial F}{\partial x} = 2x + \lambda a \longrightarrow (4) \]
\[ \frac{\partial F}{\partial y} = 2y + \lambda b \longrightarrow (5) \]
\[ \frac{\partial F}{\partial z} = 2z + \lambda c \longrightarrow (6) \]
Equating the equations to 0,
\[ \frac{\partial F}{\partial x} = 2x + \lambda a = 0 \longrightarrow (4) \]
\[ \frac{\partial F}{\partial y} = 2y + \lambda b = 0 \longrightarrow (5) \]
\[ \frac{\partial F}{\partial z} = 2z + \lambda c = 0 \longrightarrow (6) \]
\[ \Rightarrow x = \frac{-\lambda a}{2} \]
\[ \Rightarrow y = \frac{-\lambda b}{2} \]
\[ \Rightarrow z = \frac{-\lambda c}{2} \]
Substituting the values in equation (2),
\[ a\left(\frac{-\lambda a}{2} \right) + b \left(\frac{-\lambda b}{2} \right) + c\left(\frac{-\lambda c}{2} \right) = p \]
\[ \Rightarrow \lambda(a^2 + b^2 + c^2) = -2p \]
\[ \Rightarrow \lambda = \frac{-2p}{a^2 + b^2 + c^2} \]
Therefore,
\[ x = \frac{ap}{a^2 + b^2 + c^2}, y = \frac{bp}{a^2 + b^2 + c^2}, z = \frac{cp}{a^2 + b^2 + c^2} \]
The minimum value of function,
\[ f = \frac{a^2p^2}{(a^2 + b^2 + c^2)^2} + \frac{b^2p^2}{(a^2 + b^2 + c^2)^2} + \frac{c^2p^2}{(a^2 + b^2 + c^2)^2} \]
\[ = frac{p^2(a^2 + b^2 + c^2)}{(a^2 + b^2 + c^2)^2} \]
\[ = \frac{p^2}{a^2 + b^2 + c^2} \]
