Constraint minimization - Pen and paper calculation.

AIM:

To minimize the given function `f(x,y) = 5 -(x-2)^2-2(y-1)^2` and to calculate the set of parameters using the Lagrange multiplier. The following consraint given for the equation is `x + 4y = 3`.

PROBLEM STATEMENT:

1. To minimize the given equation using the concept of maxima and minima.

2. Applying the constraint equation to the given problem to find the exact values of maxima and minima.

SOLUTION PROCEDURE:

1. The equation given is as follows:

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

On simplifying the given equation we get:

`f(x,y) = 5 - (x^2 - 4cdotx +4) -2(y^2 +1 - 2y)`

`f(x,y) = -x^2 -2y^2 +4x+4y-1`_ _ _ _ (1)

2. Now we have to perform the constraint minimization using the Lagrange multiplier. Hence, we have to form a new function.

`g(x,y) = f(x,y) + lambdacdot(x+4y-3)`_ _ _ _ (2)

Substituting the equation .1 in .2 , we get :

`g(x,y) = -x^2 -2y^2+4x+4y-1 +lambdacdot(x+4y-3)`_ _ _ _ (3)

3. We need to form 3 equations using differential equation: `(delg)/(delx)=0, (delg)/(dely)=0,(delg)/(dellambda)=0`

Differenciating equation .3 with respect to x, we get the following:

`(delg)/(delx) = -2x +4+lambdacdot(1)` _ _ _ _ (4)

Differenciating equation .3 with respect to y , we get the following :

`(delg)/(dely) = -4y +4+lambdacdot(4)`_ _ _ _ (5)

Differenciating equation .3 with respect to `lambda`, we get the following:

`(delg)/(dellambda) = 0+(x+4y-3)`_ _ _ _ (6)

4. From the equations (4) , (5) ,(6):

We have to equate them to 0.

5. Equating equation (1) equal to 0 , we get:

`-2x+4+lambda=0`

`x = (4+lambda)/2`

`x=(lambda/2+2)`

6. Similarly equating equation (2) equal to 0, we get:

`-4y +4+4cdotlambda=0`

`4y = 4+4cdotlambda`

`y = 1+lambda`

7. Substituting the value of `x` and `y` in the constraint equation given we get:

`x+4y-3 =0`

`lambda/2+2+4cdot(1+lambda)-3=0`

On simplifying we get:

`lambda =-2/3`

8. Now substituting the value of `lambda` back in equation (4) and (5):

`x = (4+lambda)/2`

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

`x=5/3`

Similarly applying for equation (5):

`y =1 +lambda`

`y=1-2/3`

`y=1/3`

9. On substituting th values of `x , y `in target equation we get: `f(x,y) =5 -(5/3-2)^2 -2(1/3-1)^2`. We get `f(x,y) = 4` this is the value of minima.

RESULT:

From solving the following equations we have found `x=5/3,y=1/3,lambda=-2/3`. These are the values which minimize the target equation `f(x,y) = 5 - (x-2)^2 - 2(y-1)^2` and the constraint `x + 4y = 3`.