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To find the absolute maximum and minimum values of [tex]\( f(x, y) = x + y - xy \)[/tex] on the closed triangular region [tex]\( D \)[/tex] with vertices [tex]\((0,0)\)[/tex], [tex]\((0,2)\)[/tex], and [tex]\((6,0)\)[/tex], we need to consider the values of the function at the vertices and along the edges of the triangular region. We also need to consider any critical points inside the region.
### Step-by-Step Solution:
1. Evaluate the function at the vertices:
- At [tex]\((0, 0)\)[/tex]:
[tex]\( f(0, 0) = 0 + 0 - 0 \cdot 0 = 0 \)[/tex]
- At [tex]\((0, 2)\)[/tex]:
[tex]\( f(0, 2) = 0 + 2 - 0 \cdot 2 = 2 \)[/tex]
- At [tex]\((6, 0)\)[/tex]:
[tex]\( f(6, 0) = 6 + 0 - 6 \cdot 0 = 6 \)[/tex]
2. Evaluate the function along the edges:
- Along the edge from [tex]\((0,0)\)[/tex] to [tex]\((0,2)\)[/tex]:
- This edge is defined by [tex]\( x = 0 \)[/tex] and [tex]\( 0 \leq y \leq 2 \)[/tex].
- Substitute [tex]\( x = 0 \)[/tex] into [tex]\( f \)[/tex]: [tex]\( f(0, y) = y \)[/tex].
- So, [tex]\( f(0, y) = y \)[/tex] takes values from 0 to 2 on this edge.
- Along the edge from [tex]\((0,0)\)[/tex] to [tex]\((6,0)\)[/tex]:
- This edge is defined by [tex]\( y = 0 \)[/tex] and [tex]\( 0 \leq x \leq 6 \)[/tex].
- Substitute [tex]\( y = 0 \)[/tex] into [tex]\( f \)[/tex]: [tex]\( f(x, 0) = x \)[/tex].
- So, [tex]\( f(x, 0) = x \)[/tex] takes values from 0 to 6 on this edge.
- Along the edge from [tex]\((0,2)\)[/tex] to [tex]\((6,0)\)[/tex]:
- This edge forms a straight line connecting [tex]\((0,2)\)[/tex] and [tex]\((6,0)\)[/tex], which can be parameterized as [tex]\( y = 2 - \frac{1}{3}x \)[/tex].
- Substitute [tex]\( y = 2 - \frac{1}{3}x \)[/tex] into [tex]\( f \)[/tex]:
[tex]\[ f(x, 2 - \frac{1}{3}x) = x + (2 - \frac{1}{3}x) - x(2 - \frac{1}{3}x) = x + 2 - \frac{1}{3}x - 2x + \frac{1}{3}x^2 \][/tex]
[tex]\[ f(x, 2 - \frac{1}{3}x) = -x + 2 + \frac{1}{3}x^2 \][/tex]
- To find the extremum, take the derivative and set it to zero:
[tex]\[ \frac{d}{dx}(-x + 2 + \frac{1}{3}x^2) = -1 + \frac{2}{3}x = 0 \][/tex]
[tex]\[ x = \frac{3}{2} \][/tex]
Then, substitute [tex]\( x = \frac{3}{2} \)[/tex] back into the line equation to find [tex]\( y \)[/tex]:
[tex]\[ y = 2 - \frac{1}{3} \cdot \frac{3}{2} = 2 - \frac{1}{2} = \frac{3}{2} \][/tex]
Substitute [tex]\( x = \frac{3}{2} \)[/tex] and [tex]\( y = \frac{3}{2} \)[/tex] back into [tex]\( f \)[/tex]:
[tex]\[ f(\frac{3}{2}, \frac{3}{2}) = \frac{3}{2} + \frac{3}{2} - \frac{3}{2} \cdot \frac{3}{2} = 3 - \frac{9}{4} = \frac{3}{4} \][/tex]
- Evaluate [tex]\( f \)[/tex] at the endpoints of this edge:
- At [tex]\( (0, 2) \)[/tex], [tex]\( f(0, 2) = 2 \)[/tex].
- At [tex]\( (6, 0) \)[/tex], [tex]\( f(6, 0) = 6 \)[/tex].
3. Evaluate the critical points within the triangular region:
- To find critical points in [tex]\( D \)[/tex], compute the partial derivatives and set them to zero:
[tex]\[ f_x(x, y) = 1 - y = 0 \Rightarrow y = 1 \][/tex]
[tex]\[ f_y(x, y) = 1 - x = 0 \Rightarrow x = 1 \][/tex]
- The point [tex]\((1, 1)\)[/tex] lies inside the region [tex]\( D \)[/tex].
- At [tex]\((1, 1)\)[/tex]:
[tex]\[ f(1, 1) = 1 + 1 - 1 \cdot 1 = 1 \][/tex]
4. Summary and conclusion:
- Values of [tex]\( f \)[/tex] at vertices: [tex]\(0, 2, 6\)[/tex].
- Values of [tex]\( f \)[/tex] along edges: [tex]\(0 \text{ to } 2, 0 \text{ to } 6, \frac{3}{4} \text{ to } 6\)[/tex].
- Values at critical points inside [tex]\( D \)[/tex]: [tex]\(1\)[/tex].
Thus, the absolute maximum value is [tex]\( 6 \)[/tex] and the absolute minimum value is [tex]\( 0 \)[/tex].
### Step-by-Step Solution:
1. Evaluate the function at the vertices:
- At [tex]\((0, 0)\)[/tex]:
[tex]\( f(0, 0) = 0 + 0 - 0 \cdot 0 = 0 \)[/tex]
- At [tex]\((0, 2)\)[/tex]:
[tex]\( f(0, 2) = 0 + 2 - 0 \cdot 2 = 2 \)[/tex]
- At [tex]\((6, 0)\)[/tex]:
[tex]\( f(6, 0) = 6 + 0 - 6 \cdot 0 = 6 \)[/tex]
2. Evaluate the function along the edges:
- Along the edge from [tex]\((0,0)\)[/tex] to [tex]\((0,2)\)[/tex]:
- This edge is defined by [tex]\( x = 0 \)[/tex] and [tex]\( 0 \leq y \leq 2 \)[/tex].
- Substitute [tex]\( x = 0 \)[/tex] into [tex]\( f \)[/tex]: [tex]\( f(0, y) = y \)[/tex].
- So, [tex]\( f(0, y) = y \)[/tex] takes values from 0 to 2 on this edge.
- Along the edge from [tex]\((0,0)\)[/tex] to [tex]\((6,0)\)[/tex]:
- This edge is defined by [tex]\( y = 0 \)[/tex] and [tex]\( 0 \leq x \leq 6 \)[/tex].
- Substitute [tex]\( y = 0 \)[/tex] into [tex]\( f \)[/tex]: [tex]\( f(x, 0) = x \)[/tex].
- So, [tex]\( f(x, 0) = x \)[/tex] takes values from 0 to 6 on this edge.
- Along the edge from [tex]\((0,2)\)[/tex] to [tex]\((6,0)\)[/tex]:
- This edge forms a straight line connecting [tex]\((0,2)\)[/tex] and [tex]\((6,0)\)[/tex], which can be parameterized as [tex]\( y = 2 - \frac{1}{3}x \)[/tex].
- Substitute [tex]\( y = 2 - \frac{1}{3}x \)[/tex] into [tex]\( f \)[/tex]:
[tex]\[ f(x, 2 - \frac{1}{3}x) = x + (2 - \frac{1}{3}x) - x(2 - \frac{1}{3}x) = x + 2 - \frac{1}{3}x - 2x + \frac{1}{3}x^2 \][/tex]
[tex]\[ f(x, 2 - \frac{1}{3}x) = -x + 2 + \frac{1}{3}x^2 \][/tex]
- To find the extremum, take the derivative and set it to zero:
[tex]\[ \frac{d}{dx}(-x + 2 + \frac{1}{3}x^2) = -1 + \frac{2}{3}x = 0 \][/tex]
[tex]\[ x = \frac{3}{2} \][/tex]
Then, substitute [tex]\( x = \frac{3}{2} \)[/tex] back into the line equation to find [tex]\( y \)[/tex]:
[tex]\[ y = 2 - \frac{1}{3} \cdot \frac{3}{2} = 2 - \frac{1}{2} = \frac{3}{2} \][/tex]
Substitute [tex]\( x = \frac{3}{2} \)[/tex] and [tex]\( y = \frac{3}{2} \)[/tex] back into [tex]\( f \)[/tex]:
[tex]\[ f(\frac{3}{2}, \frac{3}{2}) = \frac{3}{2} + \frac{3}{2} - \frac{3}{2} \cdot \frac{3}{2} = 3 - \frac{9}{4} = \frac{3}{4} \][/tex]
- Evaluate [tex]\( f \)[/tex] at the endpoints of this edge:
- At [tex]\( (0, 2) \)[/tex], [tex]\( f(0, 2) = 2 \)[/tex].
- At [tex]\( (6, 0) \)[/tex], [tex]\( f(6, 0) = 6 \)[/tex].
3. Evaluate the critical points within the triangular region:
- To find critical points in [tex]\( D \)[/tex], compute the partial derivatives and set them to zero:
[tex]\[ f_x(x, y) = 1 - y = 0 \Rightarrow y = 1 \][/tex]
[tex]\[ f_y(x, y) = 1 - x = 0 \Rightarrow x = 1 \][/tex]
- The point [tex]\((1, 1)\)[/tex] lies inside the region [tex]\( D \)[/tex].
- At [tex]\((1, 1)\)[/tex]:
[tex]\[ f(1, 1) = 1 + 1 - 1 \cdot 1 = 1 \][/tex]
4. Summary and conclusion:
- Values of [tex]\( f \)[/tex] at vertices: [tex]\(0, 2, 6\)[/tex].
- Values of [tex]\( f \)[/tex] along edges: [tex]\(0 \text{ to } 2, 0 \text{ to } 6, \frac{3}{4} \text{ to } 6\)[/tex].
- Values at critical points inside [tex]\( D \)[/tex]: [tex]\(1\)[/tex].
Thus, the absolute maximum value is [tex]\( 6 \)[/tex] and the absolute minimum value is [tex]\( 0 \)[/tex].
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