PROJECT ON CONSTRAINT MINIMIZATION PROBLEM

OBJECTIVE

To minimize the following function `5-(x-2)^2 -2(y-1) ^2` subject to the following constraint `x + 4y = 3`

SOLUTION

In order to minimize the given function, we will used Lagrange multiplier method.

In mathematical optimization, the method of Lagrange multipliers is a strategy for finding the local maxima and minima of a function subject to equality constraints.

In engineering application such as in Finite element analysis, computational fluids dynamics, optimization etc. there is wide role of finding maximization and minimization of any function subjected to constraint. So here Lagrange method is most important way to optimize the function.

`g(x,y,λ)= f(x,y)+ λ.u(x,y)`

Here the function, `f(x,y)=5-(x-2)^2 -2(y-1) ^2`

And constraint function is, `u(x,y)=(x + 4y – 3)`

`g(x,y,λ) = 5-(x-2)^2 -2(y-1) ^2 + λ(x + 4y – 3)`

Condition for optimization,

`(dg)/dx = 0`

`(dg)/dy = 0`

`(dg)/(dλ ) = 0`

`(dg)/dx = -2(x-2) + λ`

`-2(x-2) + λ = 0`

`x=λ/2+ 2` ………………….equation 1

`(dg)/dy = -4(y-1) + 4λ`

`- 4(y - 1)+4λ=0`

`y= λ + 1` ………………equation 2

`(dg)/(dλ) = x + 4y – 3`

`x + 4y –3 = 0`

`x + 4y = 3` ……………… equation 3

Put the value of x and y from equation 1 and 2 in equation 3

We get

`λ/2+ 2+4(λ + 1)-3=0`

`λ/2+ 4λ +3=0`

`λ=-2/3`

Put the value of `λ` in equation 1

`x=-1/2 * 2/3+ 2`

`x=-1/3 + 2`

`x=5/3`

Put the value of `λ` in equation 2

`y= -2/3 + 1`

`y= 1/3`

So, the value of x and y are, `x=5/3` and `y= 1/3`

RESULT

At `x=5/3` and `y= 1/3` the function, `f(x,y)=5-(x-2)^2 -2(y-1) ^2` is minimized which also satisfies the constraint equation `x + 4y = 3`