Find the absolute maximum and minimum values of the function, subject to the given constraints. k(x,y)=−x2−y2 4x 4y; 0≤x≤3, y≥0,

Question

2 years ago

10

and x y≤6

1 answer:

grigory [225] · Answer 1 · 2023-12-08T22:09:18+03:00

For function k(x, y) = -x² - y² + 4x + 4y,

the absolute minimum is 0 and the absolute maximum is 6

For given question,

We have been given a function k(x, y) = -x² - y² + 4x + 4y

We need to find the absolute maximum and minimum values of the function, subject to the constraints 0 ≤ x ≤ 3, y ≥ 0, and x + y ≤ 6

First we find the partial derivative of function k(x, y) with respect to x.

⇒ $k_x=-2x+4$

Now, we find the partial derivative of function k(x, y) with respect to y.

$\Rightarrow k_y=-2y+4$

To find the critical point:

consider $k_x=0$ and $k_y=0$

⇒ -2x + 4 = 0 and -2y + 4 = 0

⇒ x = 2 and y = 2

This means, the critical point of function is (2, 2)

We have been given constraints 0 ≤ x ≤ 3, y ≥ 0, and x + y ≤ 6

Consider k(0, 0)

⇒ k(0, 0) = -0² - 0² + 4(0) + 4(0)

⇒ k(0, 0) = 0

Consider k(3, 3)

⇒ k(3, 3) = -3² - 3² + 4(3) + 4(3)

⇒ k(3, 3) = -9 - 9 + 12 + 12

⇒ k(3, 3) = -18 + 24

⇒ k(3, 3) = 6

Therefore, for function k(x, y) = -x² - y² + 4x + 4y,

the absolute minimum is 0 and the absolute maximum is 6

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