A catering business offers two sizes of baked ziti. Its small ziti dish uses 1 cup of sauce and 1 3/4 cups of cheese. Its large ziti dish uses 2 cups of sauce and 3 cups of cheese. The business has 100 cups of sauce and 400 cups of cheese on hand. It makes $6 profit on their small dishes and $5 profit on their large dishes. It wants to maximize the profit from selling the two sizes of ziti. Let x represent the number of small dishes and y represent the number of large dishes. What are the constraints for the problem?

Respuesta :

Answer:

x+2y<100

7/4x+3y<400

x>0

y>0

The constrains for the problem are:

[tex]x+2y < =100[/tex]

[tex]\frac{7x}{4} +3y < =400[/tex]

[tex]x > =0\\y > =0[/tex]

What are constraints?

Constraints are logical conditions that a solution to an optimization problem must satisfy. They reflect real-world limits on production capacity, market demand, available funds, and so on. To define a constraint, you first compute the value of interest using the decision variables.

According to question, a catering business offers two sizes of baked ziti. Its small ziti dish uses 1 cup of sauce and [tex]1\frac{3}{4}[/tex] cups of cheese. Its large ziti dish uses 2 cups of sauce and 3 cups of cheese.

The business has 100 cups of sauce and 400 cups of cheese on hand. It makes $6 profit on their small dishes and $5 profit on their large dishes.

It wants to maximize the profit from selling the two sizes of ziti.

[tex]x[/tex] represent the number of small dishes

[tex]y[/tex] represent the number of large dishes.

So, according to question, we will get following constrains:

Business has 100 cups of sauce

⇒[tex]x+2y < =100[/tex]

Business has 400 cups of cheese

⇒[tex]\frac{7x}{4} +3y < =400[/tex]

Other constrains will small and large dishes cannot be zero.

[tex]x > =0\\y > =0[/tex]

Hence we get the following constrains:

[tex]x+2y < =100[/tex]

[tex]\frac{7x}{4} +3y < =400[/tex]

[tex]x > =0\\y > =0[/tex]

Learn more about constrains here:

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