Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
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].
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
