Jonathan owns a food truck that sells tacos and burritos. He only has enough
supplies to make 90 tacos or burritos. He sells each taco for $3 and each burrito for
$7. Jonathan must sell at least $420 worth of tacos and burritos each day. If x
represents the number of tacos sold and y represents the number of burritos sold,
write and solve a system of inequalities graphically and determine one possible
solution.

When graphing inequalities, the dependent variable, or y, is placed on the left hand side of the equation while the right side has the other variables

One possible solution of the inequality is (20, 60)

The given parameters are;

The amount of supplies Jonathan has = Enough to make 90 tacos

Price at which each taco is sold = $3

Price at which each burrito is sold = $7

The amount of tacos and burritos Jonathan must sell each day = $420

Number of tacos sold = x

Number of burritos sold = y

Required:

To write and solve the system of inequalities graphically and determine one possible solution

Solution:

The system of inequalities are;

x + y ≤ 90

3·x + 7·y ≥ 420

To plot the graphs of the above inequalities, we have;

y ≤ 90 - x

y ≥ 420/7 - 3·x/7

y ≥ 60 - 3·x/7

From the graph of the inequality, the vertices of the triangle of the feasible region are;

(0, 90), (0, 60), and for accuracy, the third vertex is found as follows;

90 - x = 60 - 3·x/7

x = 52.5

y = 90 - x

∴ At the vertex point, y = 37.5

The third vertex is (52.5, 37.5)

The vertices of the triangle of the feasible region are; (0, 90), (0, 60), and (52.5, 37.5), and one possible solution inside the feasible region is (20, 60)

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jtellezd jtellezd · Answer 2 · 2021-09-30T02:15:23+02:00

To answer that question, it is necessary to formulate the two inequalities (constraints ) of the situation.

The solution is:

x = 37

y = 53

First constraint: Raw material Jonathan only has enough supplies to make either 90 tacos or 90 burritos. In other words, the number of tacos (x) plus the number of burritos (y) is at the most 90, then:

x + y ≤ 90
Second constraint: Jonathan must sell at least 420 $, then

3×x + 7×y ≥ 420

The drawing attached shows the situation and the feasible region. All the points inside that region satisfy the conditions of the problem

x + y = 90

3×x + 7×y = 420

Solving that system.

y = 90 - x

3×x + 7 × ( 90 - x ) = 420

3×x + 630 - 7×x = 420

-4×x = - 210

x = 52.5

and y = 90 - 52.5

y = 37.5

Now the solution requires integers numbers, so we look at the sells equation

3×x + 7×y ≥ 420

In this equation, each taco represent 3 $ and each burrito 7 that condition gives the idea of rounding to lower the number of tacos ( to 37 ) and rounding to bigger the number of burritos ( to 53)

Further check is

x + y ≤ 90 37 + 53 = 90 is Ok

3×x + 7×y ≥ 420

3 × ( 37) + 7 × (53) = 111 + 371 = 482 then we also meet the second constraint

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