Question

Linear Program Exercise You need to write a program that solves the following the real-world problem: A businesswoman can travel to city A, city B, or city C. It is 122 miles to city A, 237 miles to city B, and 307 miles to city C. She can travel up to 3000 miles. Dining and other expenses are $95 in city A, $130 in city B, and $180 in city C. Her expense account allows her to spend $2000. A trip to city A will generate $800 in sales, while a trip to city B will generate $1300 and a trip to city C will generate $1800. How many trips should she make to each city to maximize sales?

Answer 1

The question is a mathematical optimization problem in which a businesswoman needs to maximize sales while adhering to mileage and expense constraints. By using linear programming, with variables representing the number of trips to each city and respective constraints, she can find the optimal trip combination.

Step-by-step explanation:

The problem presented is a linear programming exercise that aims to maximize sales by determining the optimal number of trips a businesswoman should take to three different cities within her mileage and expense constraints. To address this, we can define variables x, y, and z to represent the number of trips to city A, city B, and city C, respectively.

The objective function to maximize is the total sales, given by the equation 800x + 1300y + 1800z. The constraints for this optimization problem derive from mileage and spending limits, stated as 122x + 237y + 307z ≤ 3000 for mileage, and 95x + 130y + 180z ≤ 2000 for the expense account.

Additionally, since one cannot make a negative number of trips, the variables are also subject to the non-negativity constraints x ≥ 0, y ≥ 0, and z ≥ 0. By formulating and solving this linear programming problem, possibly with graphical methods or the simplex algorithm, the businesswoman can determine the most lucrative combination of trips to make, thereby maximizing her sales within the given constraints.