Solving For P When Q Equals 4 A Detailed Mathematical Exploration

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This article delves into a mathematical problem where a quantity P exhibits a dual behavior: it is partly constant and partly varies with the square of another quantity Q. We are given two sets of values for P and Q and our mission is to determine the value of P when Q equals 4. This problem is a classic example of a combined variation problem, where a quantity depends on multiple other quantities in different ways. The approach to solving such problems involves setting up an equation that captures the given relationships, and then using the provided data points to solve for the unknown constants. Let's embark on this mathematical journey, unraveling the relationship between P and Q, and ultimately finding the value of P when Q is 4.

Setting Up the Equation: Capturing the Relationship

The cornerstone of solving this problem lies in translating the given information into a mathematical equation. The problem states that P is partly constant and partly varies as the square of Q. This implies that P can be expressed as the sum of two terms: one that remains constant regardless of the value of Q, and another that changes proportionally to the square of Q. Mathematically, this can be represented as:

P = a + bQ²

where:

  • P is the quantity we are interested in.
  • Q is the independent variable.
  • a is the constant term (the part of P that remains constant).
  • b is the constant of proportionality (determines how much P changes with the square of Q).

This equation is the heart of our solution. It encapsulates the relationship between P and Q as described in the problem. Our next step is to determine the values of the constants a and b. To do this, we will use the given data points: when Q = 2, P = 40, and when Q = 3, P = 65. By substituting these values into our equation, we will create a system of two equations with two unknowns, which we can then solve.

Leveraging the Given Data: Forming a System of Equations

Now, let's put the given data to work. We have two sets of values for P and Q:

  1. When Q = 2, P = 40
  2. When Q = 3, P = 65

We will substitute each of these sets of values into our equation P = a + bQ². This will give us two equations:

Equation 1:

40 = a + b(2)²

Simplifying, we get:

40 = a + 4b

Equation 2:

65 = a + b(3)²

Simplifying, we get:

65 = a + 9b

Now we have a system of two linear equations with two unknowns (a and b):

  • 40 = a + 4b
  • 65 = a + 9b

This system of equations can be solved using various methods, such as substitution, elimination, or matrix methods. In the next section, we will employ the elimination method to find the values of a and b. Once we have these values, we will have fully defined the relationship between P and Q and can proceed to find the value of P when Q = 4.

Solving for Constants a and b: Unveiling the Specific Relationship

We have established the following system of equations:

  1. 40 = a + 4b
  2. 65 = a + 9b

Let's use the elimination method to solve for a and b. We can eliminate a by subtracting Equation 1 from Equation 2:

(65 = a + 9b) - (40 = a + 4b)

This gives us:

25 = 5b

Dividing both sides by 5, we get:

b = 5

Now that we have the value of b, we can substitute it back into either Equation 1 or Equation 2 to solve for a. Let's use Equation 1:

40 = a + 4(5)

40 = a + 20

Subtracting 20 from both sides, we get:

a = 20

So, we have found that a = 20 and b = 5. This means our equation relating P and Q is now fully defined:

P = 20 + 5Q²

This equation tells us exactly how P changes with Q. The constant part of P is 20, and the part that varies with the square of Q is 5Q². With this equation in hand, we can finally determine the value of P when Q = 4.

Finding P When Q = 4: The Final Calculation

We have successfully determined the relationship between P and Q:

P = 20 + 5Q²

Our final task is to find the value of P when Q = 4. To do this, we simply substitute Q = 4 into our equation:

P = 20 + 5(4)²

Now, let's perform the calculation:

P = 20 + 5(16)

P = 20 + 80

P = 100

Therefore, when Q = 4, the value of P is 100.

Summarizing the Solution: A Step-by-Step Recap

Let's recap the steps we took to solve this problem:

  1. Understanding the Relationship: We recognized that P is partly constant and partly varies as the square of Q, leading us to the equation P = a + bQ².
  2. Forming a System of Equations: We used the given data points (when Q = 2, P = 40, and when Q = 3, P = 65) to create a system of two equations with two unknowns (a and b).
  3. Solving for Constants: We solved the system of equations using the elimination method, finding that a = 20 and b = 5. This gave us the specific equation P = 20 + 5Q².
  4. Calculating P: We substituted Q = 4 into our equation to find the value of P, which was 100.

This problem demonstrates a classic application of mathematical modeling. By translating a word problem into an equation and using given data to solve for unknowns, we were able to determine the value of P for a specific value of Q. This approach is widely used in various fields, including physics, engineering, and economics, to model and predict relationships between different quantities.

Real-World Applications: Beyond the Mathematical Exercise

While this problem might seem like a purely mathematical exercise, the underlying concepts have numerous real-world applications. Understanding how quantities vary with respect to each other is crucial in many fields. Here are a few examples:

  • Physics: The relationship between the distance an object falls and the time it has been falling can be modeled using a similar equation, where the distance varies with the square of the time.
  • Engineering: The load a beam can support might be partly constant (due to its own weight) and partly vary with the square of its thickness.
  • Economics: The demand for a product might have a base level (constant) and also increase with the square of the advertising expenditure.

In each of these examples, identifying the constant and variable components and establishing the mathematical relationship allows for predictions and optimizations. For instance, an engineer can use the equation relating beam thickness to load capacity to design a safe and efficient structure. Similarly, a marketing manager can use the equation relating advertising expenditure to demand to optimize their advertising budget.

Conclusion: Mastering Combined Variation Problems

This problem provided a comprehensive example of a combined variation problem, where a quantity depends on multiple factors in different ways. By breaking down the problem into smaller steps – setting up the equation, forming a system of equations, solving for constants, and performing the final calculation – we successfully found the value of P when Q = 4. The key takeaway is the ability to translate a verbal description of a relationship into a mathematical equation and then use given data to determine the specific parameters of that relationship.

Mastering these types of problems not only strengthens your mathematical skills but also provides a valuable tool for analyzing and understanding real-world phenomena. The ability to model relationships between quantities is a fundamental skill in many disciplines, making this type of problem-solving a valuable asset in various fields of study and professional endeavors.