Constraint Minimization - Pen and Paper calculation

1. AIM : To minimize a simple non-linear function Using Lagrange Multiplier, minimize a simple non-linear function `f(x,y) =5-(x-2)^2-2(y-1)^2` which is subject to the constraint `x+4y=3` .

2. INTRODUCTION :

Engineering applications are concerned with minimizing the quantities of certain engineering in various fields such as Finite Element Analysis, Computational Fluid Dynamics, etc. The purpose of these studies is to minimize a particular function w.r.t some variable with certain and then calculate optimal equation parameters.

3. BODY OF CONTENT :

Here, we have our main equation as, `f(x,y) =5-(x-2)^2-2(y-1)^2`

According to Lagrange Theory, we know that -

For a single constraint, the auxiliary function will be,

`L(x,y,λ)=f(x,y)- λg(x,y)` and

`g(x,y)=x+4y-3`

Thus,

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

Now, by doing partial differentiation ‘L’ w.r.t x, y, and to minimize the given function as,

`(∂L)/(∂x)=0`

`=>``-2x+4-λ=0` --------------- ①

`=>``x= (-λ)/2+2`

`(∂L)/(∂y)=0`

`=> -y+1-λ=0` ------------------- ②

`=> y= -λ+1`

`(∂L)/(∂λ) = 0`

`=> x+4y=3` ------------------- ③

Now, substituting the values of x and y in equation ③, we get –

`x+4y=3`

`=> (-λ)/2+2-4λ+4=3`

`=> λ= 2/3`

Now, substituting the value of in equation ①, we get –

`x= (-2)/6+2= 5/3`

Similarly, substituting the value of in equation ②, we get –

`y= (-2)/3+1= 1/3`

4. RESULT :

Therefore, for the value of `x= 5/3` , `y= 1/3` , and `λ= 2/3` , we can minimise the given function which satisfies the constraint equation.