Suppose that 2 ≤ f '(x) ≤ 5 for all values of x. What are the minimum and maximum possible values of
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For given f(x, y) the extremum: (12, 24) which is the minimum.
For given question,
We have been given a function f(x) = 4x² + 2y² under the constraint 3x+3y= 108
We use the constraint to build the constraint function,
g(x, y) = 3x + 3y
We then take all the partial derivatives which will be needed for the Lagrange multiplier equations:
Setting up the Lagrange multiplier equations:
⇒ 8x = 3λ .....................(1)
⇒ 4y = 3λ ......................(2)
constraint: 3x + 3y = 108 .......................(3)
Taking (1) / (2), (assuming λ ≠ 0)
⇒ 8x/4y = 1
⇒ 2x = y
Substitute this value of y in equation (3),
⇒ 3x + 3y = 108
⇒ 3x + 3(2x) = 108
⇒ 3x + 6x = 108
⇒ 9x = 108
⇒ x = 12
⇒ y = 2 × 12
⇒ y = 24
So, the saddle point (critical point) is (12, 24)
Now we find the value of f(12, 24)
⇒ f(12, 24) = 4(12)² + 2(24)²
⇒ f(12, 24) = 576 + 1152
⇒ f(12, 24) = 1728 ................(1)
Consider point (18,18)
At this point the value of function f(x, y) is,
⇒ f(18, 18) = 4(18)² + 2(18)²
⇒ f(18, 18) = 1296 + 648
⇒ f(18, 18) = 1944 ..............(2)
From (1) and (2),
1728 < 1944
This means, given extremum (12, 24) is minimum.
Therefore, for given f(x, y) the extremum: (12, 24) which is the minimum.
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