Find the absolute maximum and minimum values of f on the set D. f(x,y)=x^3 -3x -y^3 +12y+6, D is quadrilateral whose vertices are (-2,3), (2,3), (2,2), (-2,-2).
∂f/∂x = 3x^2-3 = 3(x^2-1)
∂f/∂y = -3y^2 + 12 = 3(4-y^2)
So any relative extrema are at (±1,±2) which are within D.
Now you just have to find extrema along the boundaries:
y=3
x=±2
y=x
To find the absolute maximum and minimum values of the function f(x, y) = x^3 - 3x - y^3 + 12y + 6 on the set D, we need to consider the critical points and boundary of the given region.
Step 1: Find the critical points
To find the critical points, we need to find the partial derivatives of f(x, y) with respect to x and y and set them equal to zero.
∂f/∂x = 3x^2 - 3 = 0
∂f/∂y = -3y^2 + 12 = 0
From the first equation, we can solve for x:
3x^2 - 3 = 0
x^2 - 1 = 0
(x - 1)(x + 1) = 0
x = 1 or x = -1
From the second equation, we can solve for y:
-3y^2 + 12 = 0
y^2 - 4 = 0
(y - 2)(y + 2) = 0
y = 2 or y = -2
So, the critical points are (1, 2), (1, -2), (-1, 2), and (-1, -2).
Step 2: Consider the boundary
The set D is a quadrilateral with the vertices (-2, 3), (2, 3), (2, 2), and (-2, -2).
We need to evaluate f(x, y) at these points.
f(-2, 3) = (-2)^3 - 3(-2) - 3^3 + 12(3) + 6 = -8 + 6 - 27 + 36 + 6 = 13
f(2, 3) = (2)^3 - 3(2) - 3^3 + 12(3) + 6 = 8 - 6 - 27 + 36 + 6 = 17
f(2, 2) = (2)^3 - 3(2) - 2^3 + 12(2) + 6 = 8 - 6 - 8 + 24 + 6 = 24
f(-2, -2) = (-2)^3 - 3(-2) - (-2)^3 + 12(-2) + 6 = -8 + 6 + 8 - 24 + 6 = -12
Step 3: Evaluate f(x, y) at critical points
We need to evaluate f(x, y) at the critical points (1, 2), (1, -2), (-1, 2), and (-1, -2).
f(1, 2) = (1)^3 - 3(1) - (2)^3 + 12(2) + 6 = 1 - 3 - 8 + 24 + 6 = 20
f(1, -2) = (1)^3 - 3(1) - (-2)^3 + 12(-2) + 6 = 1 - 3 + 8 - 24 + 6 = -12
f(-1, 2) = (-1)^3 - 3(-1) - (2)^3 + 12(2) + 6 = -1 + 3 - 8 + 24 + 6 = 24
f(-1, -2) = (-1)^3 - 3(-1) - (-2)^3 + 12(-2) + 6 = -1 + 3 + 8 - 24 + 6 = -8
Step 4: Compare the values
Now, let's compare all the values obtained:
f(-2, 3) = 13
f(2, 3) = 17
f(2, 2) = 24
f(-2, -2) = -12
f(1, 2) = 20
f(1, -2) = -12
f(-1, 2) = 24
f(-1, -2) = -8
The maximum value of f(x, y) is 24, which occurs at (2, 2) and (-1, 2).
The minimum value of f(x, y) is -12, which occurs at (-2, 3) and (1, -2).
To find the absolute maximum and minimum values of the function f(x, y) on the given set D, we need to follow these steps:
Step 1: Determine the critical points of the function f(x, y) within the interior of D.
To find the critical points, we need to find the partial derivatives of f(x, y) with respect to x and y, and set them equal to 0:
∂f/∂x = 3x^2 - 3 = 0
∂f/∂y = -3y^2 + 12 = 0
From the first equation, we get x^2 - 1 = 0, which gives x = -1 and x = 1.
From the second equation, we get y^2 - 4 = 0, which gives y = -2 and y = 2.
So, the critical points within the interior of D are (-1, -2), (-1, 2), (1, -2), and (1, 2).
Step 2: Evaluate the function at the vertices of the quadrilateral D.
The vertices of D are given as (-2, 3), (2, 3), (2, 2), and (-2, -2). Evaluate f(x, y) at each vertex:
f(-2, 3) = (-2)^3 - 3(-2) - 3^3 + 12(3) + 6
f(2, 3) = (2)^3 - 3(2) - 3^3 + 12(3) + 6
f(2, 2) = (2)^3 - 3(2) - 2^3 + 12(2) + 6
f(-2, -2) = (-2)^3 - 3(-2) - (-2)^3 + 12(-2) + 6
Step 3: Compare all the critical points and function values at the vertices to find the absolute maximum and minimum values of f(x, y).
Calculate the function value at each of the critical points:
f(-1, -2) = (-1)^3 - 3(-1) - (-2)^3 + 12(-2) + 6
f(-1, 2) = (-1)^3 - 3(-1) - 2^3 + 12(2) + 6
f(1, -2) = (1)^3 - 3(1) - (-2)^3 + 12(-2) + 6
f(1, 2) = (1)^3 - 3(1) - 2^3 + 12(2) + 6
Now, compare all the values obtained from evaluating the function at the critical points and vertices of D:
f(-2, 3) =
f(2, 3) =
f(2, 2) =
f(-2, -2) =
f(-1, -2) =
f(-1, 2) =
f(1, -2) =
f(1, 2) =
The largest value among these will be the absolute maximum, and the smallest value will be the absolute minimum.
By comparing all the function values, we find the absolute maximum and minimum values of f(x, y) on D.