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)=3y2−3x2; 2x

Question

1 year ago

13

y=9

1 answer:

Alborosie · Answer 1 · 2024-03-18T15:27:36+03:00

There is a minimum value of -81 located at (x, y) = (6, -3).

The function given to us is f(x, y) = 3y² - 3x².

The constraint given to us is 2x + y = 9.

Rearranging the constraint, we get:

2x + y = 9,

or, y = 9 - 2x.

Substituting this in the function, we get:

f(x, y) = 3y² - 3x²,

or, f(x) = 3(9 - 2x)² - 3x² = 3(81 - 36x + 4x²) - 3x² = 243 - 108x + 12x² - 3x² = 243 - 108x + 9x².

To find the extremum, we differentiate this, with respect to x, and equate that to 0.

f'(x) = - 108 + 18x ... (i)

Equating to 0, we get:

- 108 + 18x = 0,

or, 18x = 108,

or, x = 6.

Differentiating (i), with respect to x again, we get:

f''(x) = 18, which is greater than 0, showing f(x) is minimum at x = 6.

The value of y, when x = 6 is,

y = 9 - 2x,

or, y = 9 - 2*6 = 9 - 12 = -3.

The value of f(x, y) when (x, y) = (6, -3) is,

f(x, y) = 3y² - 3x²,

or, f(x, y) = 3*(-3)² - 3*6² = 3*9 - 3*36 = 27 - 108 = -81.

Thus, there is a minimum value of -81 located at (x, y) = (6, -3).

Learn more about maximum and minimum at

#SPJ4