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For today’s lunch, a school cafeteria’s budget allows it to purchase at most 60 cans of beans and 45 cans of corn. 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. If a can of beans cost $2.00 and a can of corn cost $3.00, what is the maximum amount of money required to feed all of the students either beans or corn?


A: $180


B: $195


C: $210


D: $225

Verified Answer :

I would go with B, one hundred ninety five dollars.
Hope that this would help you..

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.

\(x\leq 60\)

\(y\leq 45\)

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.

\(5x+6y\leq 420\)

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

\(x\leq 60\)

\(y\leq 45\)

\(5x+6y\leq 420\)

\(x\geq 0, y\geq 0\)       (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.

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