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beks73 [17]
1 year ago
10

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,

and x y≤6
Mathematics
1 answer:
grigory [225]1 year ago
4 0

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

Learn more about the absolute maximum and absolute minimum values of the function here:

brainly.com/question/16270755

#SPJ4

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<em> </em>

<em />

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