Nakiru, A Local Vegetable Trader In Kdtido District, Bought P P P Kilograms Of Tomatoes And Y Y Y Kilograms Of Eggplants. Tomatoes Cost UGX 2000 Per Kilogram And Eggplants Cost UGX 1000 Per Kilogram. She Had UGX 290,000 To Spend And She

Mar 11, 2025 by ADMIN 237 views

Mathematical Modeling of a Vegetable Trader's Budget

In the Kdtido District, Nakiru, a local vegetable trader, is known for her entrepreneurial spirit and ability to make the most out of her resources. Recently, she decided to invest in buying tomatoes and eggplants to sell in the local market. However, she had a limited budget of UGX 290,000 to spend on these vegetables. In this article, we will use mathematical modeling to determine the optimal amount of tomatoes and eggplants that Nakiru can buy within her budget.

Let's assume that Nakiru buys $p$ kilograms of tomatoes and $y$ kilograms of eggplants. The cost of tomatoes is UGX 2000 per kilogram, and the cost of eggplants is UGX 1000 per kilogram. Nakiru's total budget is UGX 290,000. We can represent this situation using the following equation:

2000p + 1000y ≤ 290,000

This equation represents the total cost of the tomatoes and eggplants that Nakiru can buy within her budget.

To solve this problem, we can use mathematical modeling techniques. One approach is to use linear programming, which is a method for optimizing a linear objective function subject to a set of linear constraints. In this case, our objective function is to maximize the total amount of vegetables that Nakiru can buy within her budget.

Let's define the following variables:

$p$ : the number of kilograms of tomatoes that Nakiru buys
$y$ : the number of kilograms of eggplants that Nakiru buys
$C$ : the total cost of the tomatoes and eggplants that Nakiru buys

We can represent the objective function as:

Maximize: C = 2000p + 1000y

Subject to:

2000p + 1000y ≤ 290,000

To solve this problem, we can use the graphical method or the simplex method. In this article, we will use the graphical method to find the optimal solution.

First, let's graph the constraint equation:

2000p + 1000y ≤ 290,000

We can rewrite this equation as:

2p + y ≤ 290

This is a linear inequality, and we can graph it on a coordinate plane.

Graphing the Constraint Equation

To graph the constraint equation, we can use the following steps:

Plot the line 2p + y = 290
Shade the region below the line to represent the inequality 2p + y ≤ 290

Here is the graph of the constraint equation:

  290
  ---------
  |       *
  |  *
  | *
  ---------
  0  0

Finding the Optimal Solution

To find the optimal solution, we need to find the point on the graph that maximizes the total amount of vegetables that Nakiru can buy within her budget.

Let's assume that Nakiru buys $p$ kilograms of tomatoes and $y$ kilograms of eggplants. The total cost of these vegetables is:

C = 2000p + 1000y

We want to maximize this function subject to the constraint equation:

2p + y ≤ 290

To do this, we can use the graphical method to find the point on the graph that maximizes the total cost.

The Optimal Solution

After graphing the constraint equation and finding the point on the graph that maximizes the total cost, we get the following result:

p = 145 y = 0

This means that Nakiru should buy 145 kilograms of tomatoes and 0 kilograms of eggplants to maximize the total amount of vegetables that she can buy within her budget.

In this article, we used mathematical modeling to determine the optimal amount of tomatoes and eggplants that Nakiru can buy within her budget. We represented the situation using a linear equation and used the graphical method to find the optimal solution. The result shows that Nakiru should buy 145 kilograms of tomatoes and 0 kilograms of eggplants to maximize the total amount of vegetables that she can buy within her budget.

[1] Linear Programming: An Introduction by R. J. Vanderbei
[2] Mathematical Modeling: A First Course with Modeling Applications by R. L. Borrelli and C. S. Broyden

Here is the Python code to solve the problem using the simplex method:

import numpy as np
from scipy.optimize import linprog
c = np.array([-2000, -1000])

A = np.array([[2000, 1000]])
b = np.array([290000])

x0_bounds = (0, None)
x1_bounds = (0, None)

res = linprog(c, A_ub=A, b_ub=b, bounds=[x0_bounds, x1_bounds])

print("The optimal solution is:")
print("p =", res.x[0])
print("y =", res.x[1])

This code uses the linprog function from the scipy.optimize module to solve the linear programming problem. The result is the optimal solution, which is the point on the graph that maximizes the total cost.
Mathematical Modeling of a Vegetable Trader's Budget: Q&A

In our previous article, we used mathematical modeling to determine the optimal amount of tomatoes and eggplants that Nakiru, a local vegetable trader in Kdtido District, can buy within her budget. We represented the situation using a linear equation and used the graphical method to find the optimal solution. In this article, we will answer some frequently asked questions about the problem and provide additional insights.

Q: What is the optimal solution to the problem?

A: The optimal solution to the problem is to buy 145 kilograms of tomatoes and 0 kilograms of eggplants. This means that Nakiru should prioritize buying tomatoes over eggplants to maximize the total amount of vegetables that she can buy within her budget.

Q: Why is the optimal solution to buy 145 kilograms of tomatoes and 0 kilograms of eggplants?

A: The optimal solution is to buy 145 kilograms of tomatoes and 0 kilograms of eggplants because this combination of vegetables maximizes the total cost while satisfying the constraint equation. The constraint equation is 2p + y ≤ 290, where p is the number of kilograms of tomatoes and y is the number of kilograms of eggplants. By buying 145 kilograms of tomatoes, Nakiru is able to maximize the total cost while staying within her budget.

Q: What is the total cost of the optimal solution?

A: The total cost of the optimal solution is 2000(145) + 1000(0) = 290,000. This means that Nakiru is able to buy 145 kilograms of tomatoes and 0 kilograms of eggplants within her budget of 290,000.

Q: How can Nakiru adjust her budget to buy more eggplants?

A: If Nakiru wants to buy more eggplants, she can adjust her budget by reducing the amount of tomatoes she buys. For example, if she wants to buy 100 kilograms of eggplants, she can reduce the amount of tomatoes she buys from 145 kilograms to 90 kilograms. This will allow her to stay within her budget while buying more eggplants.

Q: What are some potential limitations of the mathematical model?

A: Some potential limitations of the mathematical model include:

The model assumes that the prices of tomatoes and eggplants are fixed and do not change over time.
The model assumes that Nakiru can buy any amount of tomatoes and eggplants within her budget, but in reality, she may be limited by the availability of these vegetables.
The model does not take into account other factors that may affect Nakiru's decision, such as the quality of the vegetables, the demand for them, and the competition from other traders.

Q: How can Nakiru use the mathematical model to make better decisions?

A: Nakiru can use the mathematical model to make better decisions by:

Using the model to determine the optimal amount of tomatoes and eggplants to buy within her budget.
Adjusting her budget to buy more or fewer eggplants, depending on her priorities.
Considering other factors that may affect her decision, such as the quality of the vegetables, the demand for them, and the competition from other traders.

In this article, we answered some frequently asked questions about the mathematical model of a vegetable trader's budget. We provided additional insights into the optimal solution and discussed some potential limitations of the model. We also discussed how Nakiru can use the model to make better decisions. By using the mathematical model, Nakiru can make more informed decisions about her business and maximize her profits.

[1] Linear Programming: An Introduction by R. J. Vanderbei
[2] Mathematical Modeling: A First Course with Modeling Applications by R. L. Borrelli and C. S. Broyden

Here is the Python code to solve the problem using the simplex method:

import numpy as np
from scipy.optimize import linprog

c = np.array([-2000, -1000])

A = np.array([[2000, 1000]])
b = np.array([290000])

x0_bounds = (0, None)
x1_bounds = (0, None)

res = linprog(c, A_ub=A, b_ub=b, bounds=[x0_bounds, x1_bounds])

print("The optimal solution is:")
print("p =", res.x[0])
print("y =", res.x[1])