Beans in multiple representations

Let’s look at a table of values…

Let \(x=\) pounds of rice, \(y=\) pounds of beans.

Table

x(lbs rice) | y(lbs beans) | $ | Servings |
---|---|---|---|

0 | 0 | 0 ok | 0 no |

0 | 10 | 15 ok | 120 no |

0 | 20 | 30 ok | 240 no |

0 | 40 | 60 ok | 480 no |

0 | 50 | 75 ok | 600 ok |

0 | 60 | 90 | 720 ok |

Graph

This is getting tedious…so let’s move on to equations…,

Equations

Each lb of rice costs $2, so rice cost is \(2x\).

Each lb of beans, cost $1.5, so beans cost is \(1.5y\).

Total cost is \(2x+1.5y\) and we want that to be no more than $100.

\(2x+1.5\leq100\) dollars

Similarly, we can serve 8 people, for every lb of rice, 8x servings.

And we can serve 12 people for every lb of beans, \(12y\) servings.

We want to have at least 500 servings or

\(8x+12y\leq100\) servings

We also can’t have negative pounds or beans or rice so,

\(x\geq0, \, y\geq0\)

Let’s use desmos.com to graph these inequalities: