Question

The soccer team is selling bags of popcorn for $3 each and cups of lemonade for $2 each. To make a profit, they must collect a total of more than $120.

Write an inequality to represent the number of bags of popcorn sold, p, and the number of cups of lemonade sold, c, in order to make a profit.

Graph the solution set to the inequality on the coordinate plane

Explain how w could check if the boundary is included or excluded from the solution region

Answer 1

p = number of bags of popcorn

c = number of cups of lemonade

3p = money from just the popcorn

2c = money from just the lemonade

3p+2c = total amount of revenue

This total must be more than 120

So we arrive at the inequality 3p+2c > 120

The boundary is NOT included because there isn't an "or equal to". We exclude points on the boundary.

The boundary line 3p+2c = 120 goes through the two points (40,0) and (0,60) where p = x and c = y.

In other words, p is along the x axis and c is along the y axis.

We shade above the dashed boundary line because of the "greater than". Points in this shaded region are solution points. For instance, the point (p,c) = (30,40) is in the shaded region. This combo of popcorn and lemonade will earn more than $120 in revenue.

Points either on the dashed boundary or below it will not reach the goal of "more than $120".

See the graph below. I used Desmos to make the graph.

Answer 2

Inequality:
`\sf 3p + 2c > 120`

Explanation:

Let's denote the number of bags of popcorn sold as
`\sf p` and the number of cups of lemonade sold as
`\sf c`.

The profit,
`\sf P`, can be represented as follows:

`\sf P = 3p + 2c`

According to the given condition, the soccer team must collect more than $120 in profit, so we can write the inequality:

`\sf 3p + 2c > 120`

This is the inequality that represents the number of bags of popcorn (
`\sf p`) and the number of cups of lemonade (
`\sf c`) needed to make a profit.

To graph the solution set on the coordinate plane, we would typically rearrange the inequality into slope-intercept form (
`\sf y = mx + b`):

`\sf 3p + 2c > 120`

`\sf 2c > -3p + 120`

`\sf c > -(3)/(2)p + 60`

Now, we can graph the line
`\sf c = -(3)/(2)p + 60`,

let's find two points:

when p = 0, c = 60 so, one point is (0,60) and

When c = 0, p = 60×2 /3 = 40.

So, another point is (40,0).

Plot this point and make a dot line.

Since the inequality is
`\sf c > -(3)/(2)p + 60`, the solution region is above the line.

For Graph: See Attachment

(This forms a half-plane above the line on the coordinate plane.)