Question

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Answer 1

Answer:

The required constraints are:

$2x+2y\leq120$

$3x+4y\leq160$

$x\geq0$

$y\geq0$

Step-by-step explanation:

Consider the provided information.

Let, x is the number of roast beef specials and y is the number of turkey specials.

The profit from roast beef special is $2.30 per sandwich, the profit from turkey special is $3.10 per sandwich.

The profit function will be:

$P(x,y)=2.3x+3.10y$

The bread slices constraint will be:

The roast beef special uses two slices of bread and the turkey special uses two slices of bread. The deli has 120 slices of bread.

$2x+2y\leq120$

The cheese slices constraint will be:

The roast beef special uses three slices of cheese and the turkey special uses four slices of cheese. The deli has 160 slices of cheese.

$3x+4y\leq160$

The number of both sandwiches must be greater than zero.

Thus, the constraint will be:

$x\geq0$

$y\geq0$

Therefore, the required constraints are:

$2x+2y\leq120$

$3x+4y\leq160$

$x\geq0$

$y\geq0$

Answer 2

F (x,y) = 2.30x + 3.10y

The constraints are described with the following inequalities:
2x + 2y < = 120
3x + 4y < = 160
The corner points are: ( 0, 40) and (40/0.75, 0)
F (0,40) = 3.10 * 40 =$124
F (40/0.75,0) = (40/0.75) * 2.30 = $122.67
To maximize the profit they should sell 40 Turkey Specials.
So your answer should be C.