Question

Find the absolute maximum and minimum of f(x,y)=5x+5y with the domain x^2+y^2 less than or equal to 2^2

Suppose that f(x,y) = x^2−xy+y^2−2x+2y with D={(x,y)∣0≤y≤x≤2}

- The critical point of f(x,y)restricted to the boundary of D,
not at a corner point, is at (a,b)(. Then a=

and b= - Absolute minimum of f(x,y)
is

and absolute maximum is .

Answer #1

Find the absolute maximum and minimum of f(x,y)= 3x+4y within
the domain x^2+y^2 less than or equal tp 2^2
Find the points on the cone z^2=x^2+y^2 that are closest to the
point (3,4,0)

Find the absolute maximum and minimum vaues of f on the set
D.
f(x,y)=x^2 + y^2 + x^2y +4, D={ (x,y)| -1 less than or equal to
x less than or equal to 1, -1 less than or equal to y ess than or
equal to 1}

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 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

Calculate the following: absolute maximum, absolute minimum
values of f(x, y) = 2x^2y on -> 2x^2+y^2=6

Find the absolute maximum and minimum values of
f(x,y)=2x^2+y^2-xy^2 on the triangular region shown with vertices
(0,0), (0,4) and (4,4).

f(x)=5x^(2/3)-2x^(5/3)
a. Give the domain of f
b. Find the critical numbers of f
c. Create a number line to determine the intervals on which f is
increasing and decreasing.
d. Use the First Derivative Test to determine whether each
critical point corresponds to a relative maximum, minimum, or
neither.

Find the absolute maximum value and absolute minimum value of
f(x,y) = x2 +2y2−2x−4y+1 on the region D = {(x,y) ∈ R2 : 0 ≤ x ≤
2,|y−1.5|≤ 1.5}

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) =
4x + 6y −
x2 − y2 +
5,
D = {(x,
y) | 0 ≤ x ≤ 4, 0 ≤
y ≤ 5}
absolute maximum value
absolute minimum value

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