Answer:
Option B.
Step-by-step explanation:
Let x be the number of cans of beans and y be the number of cans of corn.
Cafeteria’s budget allows it to purchase at most 60 cans of beans and 45 cans of corn.
[tex]x\leq 60[/tex]
[tex]y\leq 45[/tex]
1 can of beans feeds 5 students, and 1 can of corn feeds 6 students. Each student will have beans or corn, but not both, and there will be a maximum of 420 students at lunch.
[tex]5x+6y\leq 420[/tex]
Can of beans cost $2.00 and a can of corn cost $3.00.
Objective function, Z=2x+3y
The required linear programming problem is
Objective function, Z=2x+3y
Subject to the constraints
[tex]x\leq 60[/tex]
[tex]y\leq 45[/tex]
[tex]5x+6y\leq 420[/tex]
[tex]x\geq 0, y\geq 0[/tex] (Only 1st quadrant)
Draw the graph of these constraints as shown below.
The verities of common shaded region are (0,45), (30,45), (60,20), (60,0), (0,0).
Points Z=2x+3y
(0,0) 0
(0,45) 135
(30,45) 195
(60,20) 180
(60,0) 120
The maximum amount of money required to feed all of the students either beans or corn is $195.
Number of cans of beans = 30
Number of cans of corn = 45
Therefore, the correct option is B.
