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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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