Yesterday I learned the beautiful method of Lagrange Multiplier, to find the minima or maxima of multivariate functions. eg
If I ask you what’s the minimum value of a function, f(x,y) = xy, subject to the constraint 5x-y = 4, the answer would be -4/5. The method is this. Ensure that first-partials of function f(x,y) exist. That is fx = (∂f)/(∂x) is not zero, and fy = (∂f)/(∂y) is not zero. Else method won’t work.
For the given example the f(x,y) = xy, because fx = y and fy = x, our method will work.
According to the method, we have 3 equations:
(1) fx + λgx = 0
(2) fy + λgy = 0
(3) 5x − y = 4
We already know what are fx and fy, right? We already evaluated them in this example; fx = y and fy = x. But what’s λ? Its called Lagrange “undetermined” multiplier. It need not be determined from above 3 equations, for determining extrema. (minima and maxima) …
Note that 3 unknowns (x, y, λ) and 3 equations (1, 2, 3) are there. Also g is the constraint function, from the constraint equation: 5x – y = 4, when g(x,y) = 0. So let’s evaluate everything to find the minima.
As we have already seen, fx = (∂f)/(∂x), so since f = xy, fx = y. Similarly fy = x. (very simple derivatives)
Now gx = (∂g)/(∂x) = (∂(5x − y − 4))/(∂x) = 5 .
Similarly gy = (∂g)/(∂y) = (∂(5x − y − 4))/(∂y) = − 1.
So 1, 2, 3 becomes:
(4) y + 5λ = 0
(5) x − λ = 0
(6) 5x − y = 4
We have 3 equations, 4, 5 and 6, and 3 unknowns, x, y, λ. Lets solve for x, y.
From 4 and 5,
(7) y + 5x = 0
and 5x − y = 4
So, 2y = − 4 or y = − 2. So x = (4 + y) ⁄ 5 = 2 ⁄ 5.
So f(x, y)minima = x*y = − 2*2 ⁄ 5 = − 4 ⁄ 5.
QED.
PS: The method can easily be extended to more variables, with x, y, z we will have 4 unknowns x, y, z, λ and 4 equations, involving them. For more than one constraint equation, we simply add over the constraint part in the above equations, that is, there will be summation sign before gx, gy; etc; ∑λgxetc.
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