CONSTRAINT MINIMIZATION PROBLEM (PEN AND PAPER TECHNIQUE)
AIM:
To solve a constraint minimization problem using Lagrange's Multiplier method
OBJECTIVE:
- To understand the concept of Lagrange's Multiplier Method
- To solve different constraint minimization problems using the above mentioned method
INTRODUCTION:
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The Lagrange multiplier technique lets you find the maximum or minimum of a multivariable function `f(x,y,...)` when there is some constraint on the input values you are allowed to use.
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This technique only applies to constraints that look something like this:
- The core idea is to find the points where the gradient vectors of `f` and `g` are parallel to each other.
- The entire process can be boiled down into setting the gradient of a certain function, called the Lagrangian, equal to the zero vector.
Process:
- Step 1: Introduce a new variable `lambda` and define a new function `mathcal{L}` as follows:
`mathcal{L}(x,y,...)` = `f(x,y,....) + lambdag(x,y,....)`
`mathcal{L}(x,y,...)` is the Lagrangian and `lambda` is the Lagrange's Multiplier
- Step 2: Solve `(delmathcal{L}(x,y,..))/(delx)`, `(delmathcal{L}(x,y,...))/(dely)` and `(delmathcal{L}(x,y,...))/(dellambda)` = 0 respectively.
- Step 3: Put the values of x and y in the original equation i.e. `f(x,y,...)` to get the minimum value.
PROBLEM:
To minimize `f(x,y) = 5 - (x-2)^2 - 2(y-1)^2` constrained to the equation `g(x,y) = x + 4y = 3`.
SOLUTION:
Given,
`f(x,y) = 5 - (x-2)^2 - 2(y-1)^2`
`g(x,y) = x + 4y - 3`
Let `lambda` be the Lagrange's Multiplier and `mathcal{L}(x,y)` be the Lagrangian.
According to the Lagrangian Multiplier method,
`mathcal{L}(x,y) = f(x,y) + lambdag(x,y)`
Now, `(delmathcal{L}(x,y))/(delx) = -2(x-2)+lambda=0` ................. `(1)`
`(delmathcal{L}(x,y))/(dely) = -4(y-1)+4lambda=0` ................... `(2)`
`(delmathcal{L}(x,y))/(dellambda) = x+4y-3=0` .................... `(3)`
Subtracting equation `(2)` from `2*(1)` we get,
`2x - y = 3`
Subtracting equation `(3)` from this new equation, we get,
`x=5y`
Putting this equation in equation `(3)` we get
`y=0.334` and `x=1.67`
So, minimum of `f(x,y)` constrained to the equation `g(x,y)` = `5-(1.67-2)^2-2(0.334-1)^2`
= `4.003988`
CONCLUSION:
Lagrange's Multiplier method is a very useful and commonly used technique for getting the constrained minimization problems. Hence, understanding this concept can help us solve numerous minimization problems with such an easy method.