Minimize Sum: Find 2 Numbers With Product 140

by Alex Johnson 46 views

Introduction

Hey there, math enthusiasts! Today, we're diving into a classic optimization problem. Optimization problems are all about finding the best possible solution – in this case, the minimum value. We’re tasked with finding two positive numbers that not only multiply to 140 but also minimize a specific sum involving these numbers. This is a fascinating blend of algebra and a touch of calculus (don't worry, we'll keep it accessible!). This concept is not just a mathematical exercise; it's a cornerstone of real-world applications, from economics to engineering, where finding optimal solutions is crucial. Understanding how to approach and solve such problems hones your analytical skills and provides a solid foundation for tackling more complex scenarios. So, let's roll up our sleeves and get started on this exciting mathematical journey!

Problem Statement

Let's break down the problem: We need to find two positive numbers. Let's call them x and y. There are two key conditions:

  1. Their product must be 140: x * y* = 140
  2. We want to minimize the sum of 7 times the first number and 5 times the second number. In other words, minimize S = 7x + 5y

This might seem a bit abstract, but think of it like this: Imagine you have a fixed area (140 square units) to build a rectangular garden. You want to minimize the cost of fencing, where one side costs 7 units per length and the other costs 5 units per length. This kind of problem pops up everywhere, making it a valuable skill to master. This step-by-step approach ensures we understand each aspect of the problem, setting us up for a clear and effective solution strategy. Identifying the constraints and the objective function is crucial in solving optimization problems.

Setting up the Equations

The beauty of math lies in its ability to translate words into equations. We've already started, but let's formalize it. We have two equations:

  1. x * y* = 140 (the constraint equation)
  2. S = 7x + 5y (the objective function – the one we want to minimize)

Our goal is to minimize S, but it currently depends on two variables, x and y. To make things easier, we need to express S in terms of just one variable. This is where the constraint equation comes in handy. We can solve the first equation for one variable (let's say y) and substitute it into the second equation. This process of reducing variables is a common technique in optimization problems. It allows us to simplify the problem and apply single-variable calculus techniques, making the solution much more manageable. By expressing the objective function in terms of a single variable, we pave the way for finding the critical points and ultimately the minimum value.

Expressing S in terms of x

From the constraint equation (x * y* = 140), we can easily solve for y:

y = 140 / x

Now, substitute this expression for y into the objective function (S = 7x + 5y):

S = 7x + 5(140 / x) S = 7x + 700 / x

Now we have S as a function of x alone. This is a significant step forward! We've transformed a two-variable problem into a single-variable one. The equation S = 7x + 700 / x is the key to unlocking our solution. It represents the sum we want to minimize in terms of just one variable, allowing us to use calculus techniques to find the minimum value. This simplification highlights the power of algebraic manipulation in solving complex problems.

Finding the Minimum Value

Now comes the calculus part. To find the minimum value of S, we'll use the following steps:

  1. Find the derivative of S with respect to x (dS/dx).
  2. Set the derivative equal to zero and solve for x. This will give us the critical points.
  3. Use the second derivative test (or analyze the first derivative) to determine if the critical point corresponds to a minimum.

Let's start with the derivative:

dS/dx = d/dx (7x + 700 / x) dS/dx = 7 - 700 / x2

Now, set the derivative to zero:

7 - 700 / x2 = 0

Solving for x:

7 = 700 / x2 x2 = 100 x = ±10

Since we are looking for positive numbers, we take x = 10. This process of finding critical points is fundamental in calculus-based optimization. By setting the derivative to zero, we identify potential locations where the function reaches a minimum or maximum. The critical point we found, x = 10, is a strong candidate for minimizing the sum, but we need to confirm it using further analysis.

Confirming the Minimum

We found a critical point at x = 10. To confirm that this corresponds to a minimum, we can use the second derivative test. This test involves finding the second derivative of S and evaluating it at the critical point.

The second derivative of S is:

d2S/dx2 = d/dx (7 - 700 / x2) d2S/dx2 = 1400 / x3

Now, evaluate the second derivative at x = 10:

d2S/dx2 (10) = 1400 / 103 = 1.4

Since the second derivative is positive (1.4 > 0) at x = 10, this confirms that we have a minimum at this point. The second derivative test is a powerful tool for determining the nature of critical points. A positive second derivative indicates a minimum, a negative second derivative indicates a maximum, and a zero second derivative requires further investigation. In our case, the positive second derivative unequivocally confirms that x = 10 leads to the smallest possible sum.

Finding the Value of y

We've found the value of x that minimizes the sum, which is x = 10. Now, we need to find the corresponding value of y. We can use the constraint equation x * y* = 140:

10 * y = 140 y = 14

So, the two positive numbers are x = 10 and y = 14. We've successfully identified the values of both variables that satisfy the given conditions. This step completes the solution process, giving us the specific numbers that minimize the sum while maintaining the product constraint. Finding the corresponding value of y is crucial for fully answering the problem and demonstrating a comprehensive understanding of the relationship between the variables.

The Solution

The two positive numbers that satisfy the conditions are:

  • x = 10
  • y = 14

Let's check if these values work:

  • Product: 10 * 14 = 140 (Correct!)
  • Sum: 7(10) + 5(14) = 70 + 70 = 140

We have successfully found the two numbers that minimize the sum while adhering to the product constraint. The final verification step is essential to ensure the accuracy of our solution. By plugging the values of x and y back into the original equations, we confirm that they satisfy both the product and the sum conditions, giving us confidence in our answer. This thorough approach reinforces the correctness of our solution and highlights the importance of double-checking results in mathematical problem-solving.

Conclusion

In this problem, we set out to find two positive numbers with a product of 140 that minimize the sum 7x + 5y. We successfully used a combination of algebra and calculus to arrive at the solution: x = 10 and y = 14. This exercise demonstrates the power of mathematical techniques in solving optimization problems. These principles are widely applicable in various fields, where finding optimal solutions is critical for efficiency and effectiveness.

Remember, the key steps were:

  1. Setting up the equations: Identifying the constraint and the objective function.
  2. Expressing the objective function in terms of one variable: Using the constraint to eliminate a variable.
  3. Finding the derivative and critical points: Using calculus to find potential minimums or maximums.
  4. Confirming the minimum: Using the second derivative test (or other methods) to verify the minimum.
  5. Finding the corresponding values: Calculating the values of all variables.

This problem showcases how mathematical concepts can be applied to real-world scenarios, fostering a deeper appreciation for the subject. Keep practicing, and you'll become a master problem-solver! If you're interested in learning more about optimization problems and calculus, I recommend checking out Khan Academy's Calculus section for comprehensive lessons and practice exercises.