Question

Find the extremum of f(x,y) subject to the given constraint, and state whether it is a maximum or a minimum. f(x,y)=4x2 2y2; 3x 3y=108

Answer 1

For given f(x, y) the extremum: (12, 24) which is the minimum.

For given question,

We have been given a function f(x) = 4x² + 2y² under the constraint 3x+3y= 108

We use the constraint to build the constraint function,

g(x, y) = 3x + 3y

We then take all the partial derivatives which will be needed for the Lagrange multiplier equations:

$f_x=8x$

$f_y=4y$

$g_x=3$

$g_y=3$

Setting up the Lagrange multiplier equations:

$f_x=\lambda g_x$

⇒ 8x = 3λ                                        .....................(1)

$f_y=\lambda g_y$

⇒ 4y = 3λ                                         ......................(2)

constraint: 3x + 3y = 108                .......................(3)

Taking (1) / (2), (assuming λ ≠ 0)

⇒ 8x/4y = 1

⇒ 2x = y

Substitute this value of y in equation (3),

⇒ 3x + 3y = 108

⇒ 3x + 3(2x) = 108

⇒ 3x + 6x = 108

⇒ 9x = 108

⇒ x = 12

⇒ y = 2 × 12

⇒ y = 24

So, the saddle point (critical point) is (12, 24)

Now we find the value of f(12, 24)

⇒ f(12, 24) = 4(12)² + 2(24)²

⇒ f(12, 24) = 576 + 1152

⇒ f(12, 24) = 1728                             ................(1)

Consider point (18,18)

At this point the value of function f(x, y) is,

⇒ f(18, 18) = 4(18)² + 2(18)²

⇒ f(18, 18) = 1296 + 648

⇒ f(18, 18) = 1944                            ..............(2)

From (1) and (2),

1728 < 1944

This means, given extremum (12, 24) is minimum.

Therefore, for given f(x, y) the extremum: (12, 24) which is the minimum.

Learn more about the extremum here:

brainly.com/question/17227640

#SPJ4