Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
To determine the vertices of the feasible region defined by the given constraints, we need to examine where the lines intersect each other and the axes.
Given constraints:
[tex]\[ \begin{cases} 2x + 3y \geq 12 \\ 5x + 2y \geq 15 \\ x \geq 0 \\ y \geq 0 \\ \end{cases} \][/tex]
1. Intersection with the y-axis:
To find where each line intersects the y-axis, set [tex]\( x = 0 \)[/tex]:
- For [tex]\( 2x + 3y = 12 \)[/tex]:
[tex]\[ 2(0) + 3y = 12 \implies 3y = 12 \implies y = 4 \][/tex]
So, the point is [tex]\( (0, 4) \)[/tex].
- For [tex]\( 5x + 2y = 15 \)[/tex]:
[tex]\[ 5(0) + 2y = 15 \implies 2y = 15 \implies y = \frac{15}{2} = 7.5 \][/tex]
So, the point is [tex]\( (0, 7.5) \)[/tex].
2. Intersection with the x-axis:
To find where each line intersects the x-axis, set [tex]\( y = 0 \)[/tex]:
- For [tex]\( 2x + 3y = 12 \)[/tex]:
[tex]\[ 2x + 3(0) = 12 \implies 2x = 12 \implies x = 6 \][/tex]
So, the point is [tex]\( (6, 0) \)[/tex].
- For [tex]\( 5x + 2y = 15 \)[/tex]:
[tex]\[ 5x + 2(0) = 15 \implies 5x = 15 \implies x = 3 \][/tex]
So, the point is [tex]\( (3, 0) \)[/tex].
3. Intersection of the two lines:
To find the intersection point of [tex]\( 2x + 3y = 12 \)[/tex] and [tex]\( 5x + 2y = 15 \)[/tex], we solve the system of equations:
[tex]\[ \begin{cases} 2x + 3y = 12 \\ 5x + 2y = 15 \\ \end{cases} \][/tex]
Multiply the first equation by 2 and the second equation by 3 to facilitate elimination:
[tex]\[ \begin{cases} 4x + 6y = 24 \\ 15x + 6y = 45 \\ \end{cases} \][/tex]
Subtract the first equation from the second:
[tex]\[ (15x + 6y) - (4x + 6y) = 45 - 24 \implies 11x = 21 \implies x = \frac{21}{11} \][/tex]
Substitute [tex]\( x = \frac{21}{11} \)[/tex] back into the first equation to solve for [tex]\( y \)[/tex]:
[tex]\[ 2\left(\frac{21}{11}\right) + 3y = 12 \implies \frac{42}{11} + 3y = 12 \implies 3y = 12 - \frac{42}{11} \implies 3y = \frac{132}{11} - \frac{42}{11} = \frac{90}{11} \implies y = \frac{30}{11} \][/tex]
So, the intersection point is [tex]\( \left(\frac{21}{11}, \frac{30}{11}\right) \)[/tex].
Finally, after determining the vertices from the constraints,
- The points are [tex]\( (0, 4) \)[/tex], [tex]\( (0, 7.5) \)[/tex], [tex]\( (3, 0) \)[/tex], and [tex]\( \left(\frac{21}{11}, \frac{30}{11}\right) \)[/tex].
Examining the given options, option (b) includes the correct points:
b) [tex]\( (0, 0), \left(0, \frac{15}{2}\right), \left(\frac{21}{11}, \frac{30}{11}\right), (6, 0) \)[/tex]
So the correct answer is:
[tex]\[ \text{b)} (0, 0), \left(0, \frac{15}{2}\right), \left(\frac{21}{11}, \frac{30}{11}\right), (6, 0) \][/tex]
Given constraints:
[tex]\[ \begin{cases} 2x + 3y \geq 12 \\ 5x + 2y \geq 15 \\ x \geq 0 \\ y \geq 0 \\ \end{cases} \][/tex]
1. Intersection with the y-axis:
To find where each line intersects the y-axis, set [tex]\( x = 0 \)[/tex]:
- For [tex]\( 2x + 3y = 12 \)[/tex]:
[tex]\[ 2(0) + 3y = 12 \implies 3y = 12 \implies y = 4 \][/tex]
So, the point is [tex]\( (0, 4) \)[/tex].
- For [tex]\( 5x + 2y = 15 \)[/tex]:
[tex]\[ 5(0) + 2y = 15 \implies 2y = 15 \implies y = \frac{15}{2} = 7.5 \][/tex]
So, the point is [tex]\( (0, 7.5) \)[/tex].
2. Intersection with the x-axis:
To find where each line intersects the x-axis, set [tex]\( y = 0 \)[/tex]:
- For [tex]\( 2x + 3y = 12 \)[/tex]:
[tex]\[ 2x + 3(0) = 12 \implies 2x = 12 \implies x = 6 \][/tex]
So, the point is [tex]\( (6, 0) \)[/tex].
- For [tex]\( 5x + 2y = 15 \)[/tex]:
[tex]\[ 5x + 2(0) = 15 \implies 5x = 15 \implies x = 3 \][/tex]
So, the point is [tex]\( (3, 0) \)[/tex].
3. Intersection of the two lines:
To find the intersection point of [tex]\( 2x + 3y = 12 \)[/tex] and [tex]\( 5x + 2y = 15 \)[/tex], we solve the system of equations:
[tex]\[ \begin{cases} 2x + 3y = 12 \\ 5x + 2y = 15 \\ \end{cases} \][/tex]
Multiply the first equation by 2 and the second equation by 3 to facilitate elimination:
[tex]\[ \begin{cases} 4x + 6y = 24 \\ 15x + 6y = 45 \\ \end{cases} \][/tex]
Subtract the first equation from the second:
[tex]\[ (15x + 6y) - (4x + 6y) = 45 - 24 \implies 11x = 21 \implies x = \frac{21}{11} \][/tex]
Substitute [tex]\( x = \frac{21}{11} \)[/tex] back into the first equation to solve for [tex]\( y \)[/tex]:
[tex]\[ 2\left(\frac{21}{11}\right) + 3y = 12 \implies \frac{42}{11} + 3y = 12 \implies 3y = 12 - \frac{42}{11} \implies 3y = \frac{132}{11} - \frac{42}{11} = \frac{90}{11} \implies y = \frac{30}{11} \][/tex]
So, the intersection point is [tex]\( \left(\frac{21}{11}, \frac{30}{11}\right) \)[/tex].
Finally, after determining the vertices from the constraints,
- The points are [tex]\( (0, 4) \)[/tex], [tex]\( (0, 7.5) \)[/tex], [tex]\( (3, 0) \)[/tex], and [tex]\( \left(\frac{21}{11}, \frac{30}{11}\right) \)[/tex].
Examining the given options, option (b) includes the correct points:
b) [tex]\( (0, 0), \left(0, \frac{15}{2}\right), \left(\frac{21}{11}, \frac{30}{11}\right), (6, 0) \)[/tex]
So the correct answer is:
[tex]\[ \text{b)} (0, 0), \left(0, \frac{15}{2}\right), \left(\frac{21}{11}, \frac{30}{11}\right), (6, 0) \][/tex]
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
