Chapter 11.7, Problem 60AYU

Mixed Nuts Nola’s Nuts, a store that specializes in selling nuts, has available 90 pounds (lb) of cashews and 120 lb of peanuts. These are to be mixed in 12-ounce (oz) packages as follows: a lower-priced package containing 8 oz of peanuts and 4 oz of cashews, and a quality package containing 6 oz of peanuts and 6 oz of cashews.

a. Using $x$ to denote the number of lower-priced packages, and use $y$ to denote the number of quality packages. Write a system of linear inequalities that describes the possible numbers of each kind of package.

b. Graph the system and label the corner points.

To determine

To find:

1. To write a system of linear inequalities that describes the possible numbers of packages of each kind of package.

2. To graph the system and label the corner points.

Answer to Problem 60AYU

Solution:

$x \ge 0, y \ge 0, 2x + 3y \le 720, 4x + 3y \le 960$

Explanation of Solution

Formula Used:

$1 oz = lb \times 16$

Given:

	Cashews	Peanuts
Lower priced packages	4oz	8oz
Quality packages	6oz	6oz

Store has $90lb$ of cashews and $120lb$ of peanuts.

Calculation:

Begin by assigning symbols for the two variables.

$x = \text{Number of lower priced packages}.$

$y = \text{Number of quality packages}.$

Because $x$ and $y$ represents the number of packages, the only meaningful values of $x$ and $y$ are non–negative.

Therefore, $x \ge 0, y \ge 0$

From the given data and the formula used we get

$4x + 6y \le 1440 => 2x + 3y \le 720$

$8x + 6y \le 1920 => 4x + 3y \le 960$

Therefore, a system of linear inequalities that describes the possible amounts of each investment are

$x \ge 0$

$y \ge 0$

$2x + 3y \le 720$

$4x + 3y \le 960$

2. The graph of the system and corner points are illustrated in the figure below.