Constraint Minimization using Lagrangian Method

Minimization of following function with respect to constraint equation

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

constraint- `lambda(x,y)=``x+4y-3`

solution:-

According to lagrangian method we consider two equations as one new equation

i.e, `g(x,y)=f(x,y) +lambda(x,y)`

for minimization partial derivatives = 0

i.e,`(partialg)/(partialx)=0`

`(partialg)/(partialy)=0`

`(partialg)/(partiallambda)=0`

now

`(partialg)/(partialx)=partial/(partialx)*(f(x,y)+lambda(x,y))`

`=partial/(partialx)*(5-(x-2)^2-2*(y-1)^2+x+4y-3)`

`=-2(x-2)+lambda`

`=-2x+4+lambda` ------->1

`(partialg)/(partialy)=partial/(partialy)*(f(x,y)+lambda(x,y))`

`=partial/(partialy)*(5-(x-2)^2-2*(y-1)^2+x+4y-3)`

`=-4(y-1)+4lambda`

`=-4y+4+4lambda` --------->2

`(partialg)/(partiallambda)=partial/(partialx)*(f(x,y)+lambda(x,y))`

`=partial/(partiallambda)*(5-(x-2)^2-2*(y-1)^2+x+4y-3)`

`=x+4y-3` -------->3

solving equations 1 and 2

eq(2)-4*eq(1)

`-4y+4+4lambda-4*(-2x+4+lambda)=0`

`-4y+4+4lambda+8x-16-4lambda=0`

`8x-4y-12=0` ----------->4

solving eq(3) and eq(4)

eq(3)+eq(4)

`x+4y-3+8x-4y-12=0`

`9x-15=0`

`x=15/9=5/3` ----------->5

substituting x value in eq(3)

we get

`5/3+4y-3=0`

`y=1/3`

substituting x value in eq(1)

`-2*5/3+4+lambda=0`

`lambda=-2/3`

finally

`therefore x=5/3,y=1/3,lambda=-2/3`

Now

`f(x,y)=5-(x-2)^2-2*(y-1)^2`

substituting the values of x,y

`f(x,y)=5-(5/3-2)^2-2*(1/3-1)^2`

`f(x,y)=4`

`therefore` The minimum value of function f(x,y) is 4.