# A confectioner sells two types of nut mixtures. The standard-mixture package contains 100 g of cashews and 200 g of peanuts and sells for \$1.85. The deluxe-mixture package contains 150 g of cashews an

A confectioner sells two types of nut mixtures. The standard-mixture package contains 100 g of cashews and 200 g of peanuts and sells for \$1.85. The deluxe-mixture package contains 150 g of cashews and 50 g of peanuts and sells for \$2.35. The confectioner has 15 kg of cashews and 20 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 100 g and the deluxe-mixture uses 150 g we must make sure that 100x + 150y ≤ 15,000 Dividing the inequality by 50 gives: 2x + 3y ≤ 300 Constraint Equation 2: There are 20 kg, or 20,000 g of peanuts available. Since the standard-mixture uses 200 g and the deluxe-mixture uses 50 g we must make sure that 200x + 50y ≤ 20,000 Dividing the inequality by 50 gives: 4x + y ≤ 400 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 ≥ y , which can be expressed as: y ≤ x MGF2106; project4R; 2021 3 Constraint Equations 4 and 5: Lastly, since she cannot produce a negative number of packages it must be that x ≥ 0 and y ≥ 0 The goal of this problem is to find the values of x and y that maximize her revenue. Since the standard-mixture sells for \$1.85 and the deluxe-mixture sells for \$2.35, her revenue will be given by the equation R = 1.85x + 2.35y If we plot the constraint equations on the same set of axes, the ordered-pair, (x,y), that maximizes her revenue will be a vertex (corner; intersection of 2 lines) of the “feasible region.” The feasible region is the graph of the constraints and can be seen here. MGF2106; project4R; 2021 4 Part 1 Exercises: 1. Find the coordinates of the vertices of the feasible region. Clearly show how the vertex is determined and which lines form the vertex. a. Find the coordinates of vertex 1. b. Find the coordinates of vertex 2. c. Find the coordinates of vertex 3. d. Find the coordinates of vertex 4. 2. Find the value of R for each vertex. a. Calculate R for vertex 1. b. Calculate R for vertex 2. c. Calculate R for vertex 3. d. Calculate R for vertex 4. 3. How many standard-mixture packages and how many deluxe-mixture packages should she sell to maximize her revenue? Part 2: Linear Programming Problem 2 Consider the feasible region in the xy-plane defined by the following linear inequalities. x ≥ 0 y ≥ 0 x ≤ 10 x + y ≥ 6 x + 2y ≤ 18 Part 2 Exercises: 1. Find the coordinates of the vertices of the feasible region. Clearly show how each vertex is determined and which lines form the vertex. 2. Find the value of Q = 58x+73y for each vertex. 3. What are the maximum and the minimum values of the function Q on the feasible region?