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Find the absolute maximum and minimum values of f on the set D. f(x, y) = x4 + y4 − 4xy + 2, D = {(x, y) | 0 ≤ x ≤ 3, 0 ≤ y ≤ 2}

Sagot :

Answer:

Absolute maximum at (3,0)

Absolute minimum at (1,1)

Step-by-step explanation:

Given - f(x, y) = x⁴ + y⁴ − 4xy + 2, D = {(x, y) | 0 ≤ x ≤ 3, 0 ≤ y ≤ 2}

To find - Find the absolute maximum and minimum values of f on the set D.

Proof -

Given that, f(x, y) = x⁴ + y⁴ − 4xy + 2

Now,

Differentiate f(x,y) with respect to x and y , we get

[tex]f_{x}(x,y)[/tex] = 4x³ - 4y

[tex]f_{y}(x,y)[/tex] = 4y³ - 4x

Now,

For the critical point, put [tex]f_{x}(x,y)[/tex] , [tex]f_{y}(x,y)[/tex]  = 0

we get

4x³ - 4x = 0

⇒x = 0, -1, 1

but given the range of x is 0 ≤x ≤ 3

⇒x = 0, 1

⇒y = 0, 1

∴ we get

2 critical points (x,y) = (0,0) , (1,1)

Now,

Put the values of x and y in f(x,y), we get

f(0,0) = 0 + 0 - 0 + 2 = 2

f(1,1) = 1 + 1 - 4 + 2 = 0

Now,

for the max and min values, finding the critical value is not enought, we have to find the values in the domain also.

Given domain is 0 ≤ x ≤ 3, 0 ≤ y ≤ 2

So, we have to check the values at the rectangle that have the coordinates

(0,0), (0,2), (3,2), (3,0)

Now,

f(0,0) = 0 + 0 - 0 + 2 = 2

f(0,2) = 0 + 16 - 0 + 2 = 18

f(3,2) = 81 + 16 - 24 + 2 = 75

f(3,0) = 81 + 0 - 0 + 2 = 83

∴ we get

Absolute maximum at (3,0)

Absolute minimum at (1,1)

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