Answer :
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]
