I Part 1: Linear Programming Problem 1 I A confectioner sells two types of nut mixtures. The standard-mixture package contains 50 g of cashews and 120 g of peanuts and sells for $1.95. The deluxe-mixture package contains 100 g of cashews and 40 g of peanuls and sells for $2.25. The confectioner has 15 kg of cashews and 18 kg of peanuts available. On the basis of past sales, the confectioner needs to have at least as many standard as deluxe packages available. How many bags of each mixture should she package to maximize her revenue? For this problem we will let x represent the number of standard-mixture packages and let y represent the number of deluxe-mixture packages. The values of x and y that will maximize her revenue depend on the amount of cashews and peanuts available for use. x = number of standard-mixture packages y = number of deluxe-mixture packages This information is used to create constraint equations because they constrain the possible values of the variables. Constraint Equation 1: There are 15 kg, or 15,000 g, of cashews available. Since the standard-mixture uses 50 g and the deluxe-mixture uses 100 9 we must make sure that 50x + 100)! S 15,000 Dividing the inequality by 50 gives: 1: + 2y 5 300 Constraint Equation 2: There are 18 kg, or 18,000 g of peanuts available. Since the standard-mixture uses 120 g and the deluxe-mixture uses 40 9 we must make sure that 120x + 40y 5 18,000 Dividing the inequality by 40 gives: 3x + y S 450 Constraint Equation 3: The problem also states that she needs to have at least as many standard as deluxe packages available so we need x 2 y , which can be expressed as: y 5 x