Question

What are the absolute maximum and minimum values of the function

f(x, y) = (x+ 3)^2+ (y−1)^2 on the region enclosed by the circle (x−1)^2+ (y+ 2)^2= 4?

Answer #1

Find the absolute maximum, and minimum values of the function:
f(x, y) = x + y − xy Defined over the closed rectangular region D
with vertices (0,0), (4,0), (4,2), and (0,2)

find the absolute maximum value and absolute minimum values of
the function f(x,y)4xy^2-x^2y^2-xy^3 on the set D, where D is the
closed trianglar region in the xy-plane with certices
(0,0)(0,6)(6,)0

Find the absolute maximum and absolute minimum values of f(x,y)
= x^2 + 2y^2 − 2x + 2 on the closed disk D: x^2 + y^2 ≤ 4.
Answer: absolute min: f(1, 0) = 1; absolute max: f(−1, ± √3) =
11

Find the absolute maximum value and the absolute minimum value
of the function f ( x , y ) = x 2 y 2 + 3 y on the set D defined as
the closed triangular region with vertices ( 0 , 0 ), ( 1 , 0 ),
and ( 1 , 1 ), that is, the set D = { ( x , y ) | 0 ≤ x ≤ 1 , 0 ≤ y
≤ x }...

Find the absolute maximum and minimum values of f on
the set D.
f(x, y) =
4x + 6y −
x2 − y2 +
3,
D = {(x,
y) | 0 ≤ x ≤ 4, 0 ≤
y ≤ 5}
absolute maximum value
absolute minimum value

Find the absolute maximum and minimum values of f on
the set D.
f(x, y) =
x2 + y2 +
x2y + 6,
D = {(x,
y) | |x| ≤ 1,
|y| ≤ 1}
absolute maximum value
absolute minimum value

Find the absolute maximum and minimum values of f on
the set D.
f(x, y) =
4x + 6y −
x2 − y2 +
5,
D = {(x,
y) | 0 ≤ x ≤ 4, 0 ≤
y ≤ 5}
absolute maximum value
absolute minimum value

Let f(x) = x 1/2 (3−x). Find the absolute maximum and absolute
minimum values of the f(x) on the interval [1, 3].

Find the absolute maximum and minimum values of f on
the set D.
f(x, y) =
4x + 6y −
x2 − y2 +
2,
D = {(x,
y) | 0 ≤ x ≤ 4, 0 ≤
y ≤ 5}

dentify the absolute minimum and absolute maximum values of the
function f(x)=x3−12x2−27x+22 f ( x ) = x 3 - 12 x 2 - 27 x + 22 on
the interval [−2,4]

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