At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
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].
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
