Question

Let f(x, y)=2x^2-3y^2-x+y+1, find the maximum and minimum values of f(x, y) over the triangle with vertices (0,0),(1,0), and (0,1), shown below:

Answer 1

The maximum value of f(x, y) over the given triangle is 2 and the minimum value is -2.

Step-by-step explanation:

To find the maximum and minimum values of the function f(x, y) = 2x^2 - 3y^2 - x + y + 1 over the triangle with vertices (0,0), (1,0), and (0,1), we can evaluate the function at each vertex and check for additional critical points.

Substituting the vertex coordinates into f(x, y), we get:

• f(0, 0) = 2(0)^2 - 3(0)^2 - 0 + 0 + 1 = 1
• f(1, 0) = 2(1)^2 - 3(0)^2 - 1 + 0 + 1 = 2
• f(0, 1) = 2(0)^2 - 3(1)^2 - 0 + 1 + 1 = -2

Therefore, the maximum value of f(x, y) over the triangle is 2 and the minimum value is -2.