Consider the function f(x, y) = 4x2 搰 3y2 2xy over the unit square 0...
Maximum and Minimum Values of f on each Edge of the Square:
To find the maximum and minimum values of the function f(x, y) on each edge of the square, we need to substitute the values of x and y that lie on each edge into the function and evaluate the function.
Edge 1: x = 0, 0 ≤ y ≤ 1
Substituting x = 0 into the function f(x, y) = 4x^2 - 3y^2 + 2xy, we get:
f(0, y) = -3y^2
The maximum value of f(0, y) occurs when y = 0, which gives f(0, 0) = 0.
The minimum value of f(0, y) occurs when y = 1, which gives f(0, 1) = -3.
Edge 2: x = 1, 0 ≤ y ≤ 1
Substituting x = 1 into the function f(x, y) = 4x^2 - 3y^2 + 2xy, we get:
f(1, y) = 4 - 3y^2 + 2y
To find the maximum and minimum values of f(1, y), we can take the derivative of f(1, y) with respect to y and set it equal to zero:
df(1, y)/dy = -6y + 2 = 0
Solving for y, we get y = 1/3.
Substituting y = 1/3 into f(1, y), we get:
f(1, 1/3) = 4 - 3(1/3)^2 + 2(1/3) = 4 - 1 + 2/3 = 11/3
The maximum value of f(1, y) occurs when y = 1/3, which gives f(1, 1/3) = 11/3.
The minimum value of f(1, y) occurs when y = 0 or y = 1, which gives f(1, 0) = 4 and f(1, 1) = 3.
Edge 3: y = 0, 0 ≤ x ≤ 1
Substituting y = 0 into the function f(x, y) = 4x^2 - 3y^2 + 2xy, we get:
f(x, 0) = 4x^2
The maximum value of f(x, 0) occurs when x = 1, which gives f(1, 0) = 4.
The minimum value of f(x, 0) occurs when x = 0, which gives f(0, 0) = 0.
Edge 4: y = 1, 0 ≤ x ≤ 1
Substituting y = 1 into the function f(x, y) = 4x^2 - 3y^2 + 2xy, we get:
f(x, 1) = 4x^2 - 3 + 2x
To find the maximum and minimum values of f(x,