Constraint Minimization - Pen and Paper calculation

Aim: In this project, we have to minimize a simple, non-linear function using Lagrange Multiplier. The function is `f(x,y)=5-(x-2)^(2)-2(y-1)²` which is subject to the following constraint `x+4y=3`.

Introduction:

Engineering applications are concerned with maximizing or 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 constraints and then calculate optimal equation parameters.

Solution:

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

According to Lagrange Theory, wkt

For a single constraint, the auxiliary function will be

`L(x,y,λ)=f(x,y)-λg(x,y)`

Now, we perform constraint minimization, we get;

`f(x,y)=5-(x-2)^2-2(y-1)²`

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

Thus,

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

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

Now, by partially differentiating ‘L’ w.r.t x, y and λ to minimise the given function.

`∂L/∂x=0`

`-2x+4-λ=0` ------> considering this as equation 1

`x=(-λ)/2+2`

`∂L/∂y=0`

`-y+1-λ=0` -------> considering this as equation 2

`y=-λ+1`

`∂L/∂λ=0`

`x+4y=3` ------> considering this as equation 3

Substituting the values of x and y in equation 3, we get

`x+4y=3`

`(-λ)/2+2-4λ+4=3`

Therefore,

`λ=2/3`

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

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

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

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

RESULT:

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