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.


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