Problem 53 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Correspond to those in Exercises \(15-27\), Section \(5.1 .\) Use the results of your previous work to help you solve these problems. The Coffee Shoppe sells a coffee blend made from two coffees, one costing $$\$ 5 / \mathrm{lb}$$ and the other costing $$\$ 6 / \mathrm{lb}$$. If the blended coffee sells for $$\$ 5.60 / \mathrm{lb}$$, find how much of each coffee is used to obtain the desired blend. Assume that the weight of the blended coffee is $$100 \mathrm{lb}$$.

Step-by-Step Solution

Answer

The coffee blend is made from 40 lb of \(\$ 5 / \mathrm{lb}\) coffee and 60 lb of \(\$ 6 / \mathrm{lb}\) coffee.

Step 1: Define the variables

Let x be the amount (in lb) of the $$\$ 5 / \mathrm{lb}$$ coffee and y be the amount (in lb) of the $$\$ 6 / \mathrm{lb}$$ coffee. The total weight of the coffee blend is 100 lb. Therefore, we have the following equation representing the total weight: \(x+y=100\) Now, we construct an equation for the total cost of the blend. If the blended coffee sells for $$\$ 5.60 / \mathrm{lb},$$ then the total cost of the 100 lb blend is \(100*5.60\). The total cost can also be represented as: $$5x+6y=5.60(100)$$ #Step 2: Solve the System of Equations#

Step 2: Elimination method

We will use the elimination method to solve the system. Firstly, we will multiply the first equation (the weight equation) by -5 to eliminate x in both equations. The equations now look like this: \(-5(x+y)=-5(100)\) Multiply to get: \(-5x-5y=-500\) Now we have: \(-5x-5y=-500\) \(5x+6y=560\) Next, add these two equations to eliminate x: \((-5x - 5y) + (5x + 6y)=-500 + 560\) Simplify the equation: \(y=60\) #Step 3: Substitute and Solve#

Step 3: Substitute the value of y

Now that we have the value of y, we can substitute it into the weight equation to find the value of x: \(x + 60 = 100\) Subtract 60 from both sides: \(x = 100 - 60\) So, \(x = 40\) #Step 4: Interpret the Solution#

Step 4: Determine the amount of each type of coffee

We have determined the values for x and y: \(x = 40\) and \(y = 60\). Therefore, the coffee blend is made from 40 lb of $$\$ 5 / \mathrm{lb}$$ coffee and 60 lb of $$\$ 6 / \mathrm{lb}$$ coffee.

Key Concepts

Algebraic Methods

Algebraic methods are fundamental techniques in mathematics used to solve various types of equations, including linear, quadratic, and polynomial equations. They are the backbone of algebra and provide a systematic way to find unknown values that satisfy given conditions. Students often begin learning algebraic methods with simple operations such as addition, subtraction, multiplication, and division and then move on to more complex concepts like factoring, expanding, and working with exponents.
In the context of our coffee blend problem from the textbook, algebraic methods come into play when we define variables to represent the quantities we are trying to find, setting up equations based on the information provided. The equations created are representative of the relationships between the quantities, prices, and weights in this scenario. Understanding how to manipulate and solve these equations is pivotal to determining the correct amount of each type of coffee needed for the blend.

Elimination Method

The elimination method is a powerful algebraic technique used to solve systems of linear equations, where we seek to find the value of multiple variables that make the equations true simultaneously. This method works by eliminating one of the variables in order to reduce the system to a single equation with one variable, which can then be solved with basic algebra.
Here's how it applies to our coffee shop problem: We use the elimination method by first equalizing coefficients for one of the variables across both equations. In our case, multiplying the total weight equation by -5 allows us to cancel out the variable 'x' when we add it to the cost equation, leaving us with an equation that only contains 'y'. Once we find the value of 'y', we can trace our steps back to find the value of 'x'. This step-by-step approach simplifies complex problems making them much more manageable.

Applied Mathematics

Applied mathematics involves the use of mathematical methods and theories to solve practical problems from various disciplines such as science, engineering, business, and economics. It is in the application of these methods that students can see the real-world value of the mathematical concepts they learn in the classroom.
For instance, in our coffee shop example, applied mathematics is utilized to figure out how to mix different types of coffee to achieve a desired blend. The abstract equations we construct are not just numbers and symbols; they represent actual quantities of coffee and money. By finding the solution to these equations, we provide tangible information that a coffee shop owner can use to make informed financial and production decisions. Applied mathematics turns theoretical knowledge into tools that can tackle everyday challenges, bridging the gap between abstraction and reality.