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) = x^2 + 4y^2-3xy; x + y = 16 value of Q located at (x, y) = (Simplify your answers.) There is a

Answer 1

Answer:

  (x, y) = (11, 5)

  f(x, y) = 56, a minimum

Step-by-step explanation:

You apparently want the location and value of the extremum of f(x, y) = x² +4y² -3xy, subject to the constraint x + y = 16.

Objective function

Applying the constraint to write y in terms of x, the function can be expressed in terms of a single variable as ...

  f(x, y) = f(x, 16 -x) = x² +4(16 -x)² -3x(16 -x)

  f(x) = x² +4(256 -32x +x²) -48x +3x² = 8x² -176x +1024

We can write this in vertex form to find the extreme value.

  f(x) = 8(x² -22x +128) = 8((x -11)² +7)

  f(x) = 8(x -11)² +56 . . . . . . . . . . a minimum of 56 at x = 11

  y = 16 -x = 16 -11 = 5

The minimum value is 56 at (x, y) = (11, 5).

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Additional comment

You get the same result using the method of Lagrange multipliers.

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