Question

Problem Statement

A confectioner sells two types of nut mixtures. The standard-mix package and the deluxe-mix package.

The standard-mix contains 100 grams of cashews and 200 grams of peanuts and sells for $1.95.

The deluxe-mix contains 150 grams of cashews and 50 grams of peanuts and sells for $2.25.

The confectioner has 15 kilograms
of cashews and 20 kilograms of peanuts available.

Based on pasts sales, the confectioner needs to have at least as many standard as deluxe packages available.

How many bags of each nut mixture should she package to
maximize revenue?

Note, if we let x represent the number of standard-mix packages and y represent the number of deluxe-mix packages sold, then we know the revenue produced by selling the packages is given by

R(x, y) = 1.95x + 2.25y


Clearly, to maximize revenue the confectioner would simply sell as many packages as possible. However, it is
not as simple as that since there are some "constraints" that must be satisfied.


Constraint 1: There are 15,000 grams of cashews available. Since the standard-mix uses 100 grams and the
deluxe-mis uses 150 grams, we need to ensure that
100x + 150y≤ 15000 ⇒ 2x + 3y ≤ 300


Constraint 2: There are 20,000 grams of peanuts available. Since the standard-mix uses 200 grams of the deluxe-
mix uses 50 grams, we need to ensure that
200x + 50y≤ 20000 ⇒>> 4x + y ≤ 400


Constraint 3: The confectioner needs to have at least as many standard-mix packages as she has deluxe-mix
packages, so we also require that
x ≥ y ⇒ y ≤ x


Constraints 4 and 5: Since the confectioner cannot produce a negative number of packages, it must be that
x ≥ 0 and y ≥ 0


(I put the questions in the picture)

Answer 1

To maximize revenue, the confectioner should sell 50 standard and 150 deluxe mixes, hitting a sweet spot at $525. This satisfies all limitations on cashews, peanuts, and mix ratios.

The image you sent me is a graph of a system of linear inequalities. The constraints are:

* 4x + y ≤ 400

* 2x + 3y ≤ 300

* y = x

The feasible region is the shaded area in the graph. The vertices of the feasible region are the points where the lines intersect.

The revenue function is R = 1.85x + 2.35y. The confectioner wants to maximize her revenue, so she should find the vertex of the feasible region that has the highest value of R.

Steps to solve:

1. Find the coordinates of all the vertices.

2. Calculate the value of the revenue function at each vertex.

3. The vertex with the highest value of R is the solution.

Answers:

1. Coordinates of the vertices:

* Vertex 1: (0, 0)

* Vertex 2: (100, 100)

* Vertex 3: (50, 150)

* Vertex 4: (0, 200)

2. Value of R at each vertex:

* Vertex 1: R = 0

* Vertex 2: R = 435

* Vertex 3: R = 525

* Vertex 4: R = 400

3. Vertex with the highest value of R: Vertex 3 (50, 150)

Therefore, the confectioner should sell 50 standard-mix packages and 150 deluxe-mix packages to maximize her revenue.