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f(x,y)= 2-3x+2y , D the closed region enclosed by the triangle vertices (0,0), (4,0), (0,6). Find the absolute extreme values. Please thourouly explain each step.

User Fedj
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Answer:

To find the absolute extreme values of a function on a closed, bounded region, we need to follow the following steps:

1.Find the critical points of the function in the interior of the region (i.e., points where the partial derivatives are equal to zero).

2.Evaluate the function at these critical points.

3.Evaluate the function at the vertices of the region.

4.Compare the values obtained in steps 2 and 3 to find the absolute extreme values.

Let's follow these steps for the given function f(x,y) = 2-3x+2y, over the region D, enclosed by the triangle vertices (0,0), (4,0), (0,6).

Step 1: Find the critical points in the interior of the region

To find the critical points, we need to find where the partial derivatives of f(x,y) are equal to zero:

∂f/∂x = -3 = 0, and

∂f/∂y = 2 = 0.

These partial derivatives are never zero simultaneously, so there are no critical points in the interior of the region.

Step 2: Evaluate the function at the vertices of the region

The vertices of the region are (0,0), (4,0), and (0,6). We need to evaluate the function at each of these points:

f(0,0) = 2

f(4,0) = -10

f(0,6) = 10

Step 3: Compare the values obtained in steps 2 and 3 to find the absolute extreme values

The minimum value of f(x,y) over the region D is -10, which occurs at the point (4,0). The maximum value of f(x,y) over the region D is 10, which occurs at the point (0,6).

Therefore, the absolute extreme values of f(x,y) over the region D are:

Minimum value: -10, at (4,0)

Maximum value: 10, at (0,6)

So, the absolute extreme values are -10 and 10, at the points (4,0) and (0,6), respectively.

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User Sergey Kornilov
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