Problem 19
Question
The coffee variety Arabica yields about \(750 \mathrm{~kg}\) of coffee beans per hectare, while Robusta yields about 1200 \(\mathrm{kg} /\) hectare. In Problems \(19-20\), suppose that a plantation has a hectares of Arabica and \(r\) hectares of Robusta. \(^{8}\) Write an equation relating \(a\) and \(r\) if the plantation yields \(1,000,000 \mathrm{~kg}\) of coffee.
Step-by-Step Solution
Verified Answer
Question: Find the equation that relates the hectares of Arabica (a) and hectares of Robusta (r) based on their yields and the given total production.
Answer: The equation that relates the hectares of Arabica and Robusta is 5a + 8r = 6666.67.
1Step 1: Determine the yield of each variety
We first need to determine how much coffee is produced from each variety.
- For Arabica, we multiply the hectares (a) by the yield per hectare of Arabica (750 kg):
750a
- For Robusta, we multiply the hectares (r) by the yield per hectare of Robusta (1200 kg):
1200r
2Step 2: Combine the yields to match the total production
Now that we have the amounts produced by both Arabica and Robusta, we will add them together, since this sum should equal the total production of 1,000,000 kg. Therefore, we get the equation:
750a + 1200r = 1,000,000
3Step 3: Simplify the equation (if possible)
The given equation is 750a + 1200r = 1,000,000, and we can see that all the coefficients are divisible by 150. Dividing all terms of the equation by 150 will make the equation easier to work with:
(750a)/150 + (1200r)/150 = (1,000,000)/150
5a + 8r = 6666.67
So, the simplified equation relating the hectares of Arabica and Robusta is:
5a + 8r = 6666.67
Key Concepts
Understanding Linear EquationsApplying Problem Solving TechniquesExploring Agricultural Mathematics
Understanding Linear Equations
Linear equations are equations of the first order. An equation like this represents a straight line when plotted on a graph. The general form of a linear equation is \[ ax + by = c \] where:
Linear equations like this are fundamental in algebra because they are the basis for analyzing and solving many real-world problems.
By learning how to manipulate and solve these equations, you develop skills that allow you to work with more complex algebraic structures. Remember, the key is in balancing the equation so that both sides remain equal.
- \(a\), \(b\), and \(c\) are constants.
- \(x\) and \(y\) are variables.
Linear equations like this are fundamental in algebra because they are the basis for analyzing and solving many real-world problems.
By learning how to manipulate and solve these equations, you develop skills that allow you to work with more complex algebraic structures. Remember, the key is in balancing the equation so that both sides remain equal.
Applying Problem Solving Techniques
Problem solving in mathematics often involves certain strategies, such as identifying what information is given and what needs to be found. In our exercise:
This process involves translating the problem into a mathematical form, which in this case is a linear equation.
First, identify the yield from each type of coffee plant:
Next, combine them to form an equation that sums up to the total yield: \[ 750a + 1200r = 1,000,000 \]
Simplifying this makes it easier to interpret and solve, as shown in our solution with the equation \[ 5a + 8r = 6666.67 \].
Throughout this process, clarity in each step ensures the problem is efficiently solved.
- We have been given the yield per hectare for Arabica and Robusta.
- We need to find a relationship between the hectares planted of each to achieve a total yield of 1,000,000 kg.
This process involves translating the problem into a mathematical form, which in this case is a linear equation.
First, identify the yield from each type of coffee plant:
- Arabica: \(750a\)
- Robusta: \(1200r\)
Next, combine them to form an equation that sums up to the total yield: \[ 750a + 1200r = 1,000,000 \]
Simplifying this makes it easier to interpret and solve, as shown in our solution with the equation \[ 5a + 8r = 6666.67 \].
Throughout this process, clarity in each step ensures the problem is efficiently solved.
Exploring Agricultural Mathematics
Agricultural mathematics involves using mathematical models to solve problems related to agriculture. These problems can involve optimizing resources like land or predicting the outcomes based on different farming strategies.
In the context of our exercise, we are optimizing land allocation for two types of coffee plants to meet a specific production goal.
This exercise reflects real agricultural decisions where mathematical calculations help to maximize production while minimizing costs and ensuring sustainable farming practices.
Understanding concepts like linear equations aids in making such decisions, providing a clearer picture of how changes in one area (e.g., the number of hectares allocated) can affect overall outcomes.
In the context of our exercise, we are optimizing land allocation for two types of coffee plants to meet a specific production goal.
- Determining how many hectares should be dedicated to Arabica and Robusta to achieve a yield of 1,000,000 kg.
- Using equations to model potential yields because resources are finite and must be efficiently used.
This exercise reflects real agricultural decisions where mathematical calculations help to maximize production while minimizing costs and ensuring sustainable farming practices.
Understanding concepts like linear equations aids in making such decisions, providing a clearer picture of how changes in one area (e.g., the number of hectares allocated) can affect overall outcomes.
Other exercises in this chapter
Problem 18
Identify the initial value and the rate of change, and explain their meanings in practical terms. The value of an antique is \(2500+80 n\) dollars, where \(n\)
View solution Problem 18
Rewrite the function in slope-intercept form. $$ j(t)=1.2+0.4(t-5) $$
View solution Problem 19
Solve the systems of equations. $$ \left\\{\begin{array}{l} 3 e+2 f=4 \\ 4 e+5 f=-11 \end{array}\right. $$
View solution Problem 19
Write an equation in point-slope form for the line. Through (6,-7) and (-6,-1)
View solution