Find two numbers whose difference is 42 and whose product is minimized. Step 1: If two numbers have a difference of 42, how can we express these numbers in terms of a variable?

Let the two numbers be x and y, with x > y. Since their difference is 42, we can express y as y = x - 42.

Given the two numbers x and y with y = x - 42, how do we express their product?

The product P of the two numbers is P = x y. Substituting y = x - 42, we get P = x (x - 42) = x^2 - 42x.

How do we find the value of x that minimizes the product P = x^2 - 42x?

To minimize P, we take the derivative of P with respect to x, set it to zero, and solve for x. So, dP/dx = 2x - 42 = 0.

What is the critical point for the quadratic function P = x^2 - 42x, and how do we find it?

Setting the derivative 2x - 42 = 0 gives x = 21. This is the critical point where P may be minimized.

What are the corresponding values of the two numbers when x = 21?

Since y = x - 42, when x = 21, y = 21 - 42 = -21. The two numbers are 21 and -21.

What is the minimum product of the two numbers whose difference is 42?

The minimum product is P = 21 (-21) = -441.

Why is the product minimized when the two numbers are 21 and -21?

Because the quadratic P = x^2 - 42x opens upwards, its vertex at x = 21 gives the minimum value of P, corresponding to the pair (21, -21).

Can the two numbers be both positive or both negative to achieve the minimum product?

No, to minimize the product given a fixed difference of 42, one number must be positive and the other negative, as in 21 and -21, which yields the most negative (minimum) product.

What is the significance of the sign of the product in this problem?

A negative product indicates the numbers have opposite signs, which is necessary to achieve the minimum product when the difference is fixed at 42.

Is the pair (21, -21) unique for this problem?

Yes, the pair (21, -21) uniquely minimizes the product for the given difference of 42, as shown by the vertex of the quadratic function.

Find Two Numbers Whose Difference Is 42 And Whose Product Is A Minimum. Step 1 If Two Numbers Have A

Find Two Numbers Whose Difference Is 42 And Whose Product Is A Minimum. Step 1 If Two Numbers Have A

When faced with a problem involving two numbers with a specific difference and a goal to minimize or maximize their product, it introduces an interesting challenge in algebra and optimization. In this article, we will explore the process of finding two numbers whose difference is 42 and whose product is minimized. This type of problem is common in various fields such as mathematics, engineering, and economics, where optimization plays a crucial role. We will go through step-by-step methods, including setting up equations, applying calculus, and analyzing the results to achieve an optimal solution.

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Understanding the Problem: Finding Two Numbers with a Given Difference and Minimum Product

Before diving into complex calculations, it's essential to understand what the problem entails:

Given Condition: The difference between two numbers is 42.

Objective: Find the two numbers such that their product is as small as possible.

Constraints: The difference between the two numbers remains fixed at 42.

This problem can be approached systematically by defining variables, forming algebraic expressions, and then applying calculus to find the minimum of the resulting function.

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Step 1: Define Variables and Set Up the Equation

The first step in solving the problem involves choosing appropriate variables.

Choosing Variables

Let the two numbers be \( x \) and \( y \).
Since the difference between the numbers is 42, we can represent this as:

\[ y - x = 42 \]

To simplify, assume \( x \) is one number, then \( y = x + 42 \).

Expressing the Product

The goal is to minimize the product \( P \) of these two numbers:

\[
P = x \times y
\]

Substitute \( y \) with \( x + 42 \):

\[
P(x) = x \times (x + 42) = x^2 + 42x
\]

Our task reduces to finding the value of \( x \) that minimizes \( P(x) \).

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Step 2: Find the Critical Points Using Calculus

To determine the minimum value of the product, calculus offers powerful tools.

Differentiate the Product Function

Compute the first derivative of \( P(x) \):

\[ P'(x) = 2x + 42 \]

Set \( P'(x) = 0 \) to find critical points:

\[ 2x + 42 = 0 \] \[ 2x = -42 \] \[ x = -21 \]

Find Corresponding \( y \)

Recall that \( y = x + 42 \):

\[ y = -21 + 42 = 21 \]

Thus, the two numbers are \( x = -21 \) and \( y = 21 \).

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Step 3: Verify the Nature of Critical Points

To confirm whether this critical point corresponds to a minimum, examine the second derivative.

Second Derivative Test

The second derivative of \( P(x) \):

\[ P''(x) = 2 \]

Since \( P''(x) = 2 > 0 \), the function \( P(x) \) is concave upward at \( x = -21 \), indicating a minimum point.

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Step 4: Calculate the Minimum Product

Having identified \( x = -21 \), compute the corresponding product:

\[
P_{\text{min}} = x \times y = (-21) \times 21 = -441
\]

The minimum product of the two numbers with a difference of 42 is -441.

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Step 5: Interpret the Results

The two numbers are -21 and 21.
Their difference:

\[ 21 - (-21) = 42 \]

Their product:

\[ -21 \times 21 = -441 \]

This result indicates that the minimal product occurs when the two numbers are symmetric around zero, with one being negative and the other positive.

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Additional Insights and Variations

Understanding this problem allows for exploration of related scenarios:

1. When Both Numbers Are Positive

If both numbers are positive and their difference is 42, then:

\[ x > 0, \quad y = x + 42 > 0 \]

The product:

\[ P = x(x + 42) \]

Since the parabola opens upward, the minimum occurs at:

\[ x = -\frac{b}{2a} = -\frac{42}{2} = -21 \]

But \( x \) must be positive, so the minimum in the positive domain occurs at the smallest possible \( x \) satisfying the condition, which is at \( x = 0 \).
At \( x = 0 \), \( y = 42 \), and the product:

\[ P = 0 \times 42 = 0 \]

Therefore, the minimal positive product with both numbers positive is 0 at \( (0, 42) \).

2. Comparing Negative and Positive Solutions

The previous analysis shows that allowing negative numbers yields a more extreme minimum of -441, while restricting to positive numbers yields a minimal product of 0.

3. Real-world Applications

This type of problem is useful in fields like physics for minimizing energy configurations, economics for profit/loss optimization, and engineering for minimizing costs under constraints.

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Conclusion: How to Find Two Numbers with a Fixed Difference and a Minimized Product

In summary, solving for two numbers with a fixed difference to minimize their product involves:

Setting variables and expressing one in terms of the other.

Formulating the product as a function of a single variable.

Applying calculus (derivatives) to find critical points.

Using the second derivative test to confirm minimum points.

Interpreting the results to identify the specific numbers.

This process is a fundamental application of algebra and calculus in solving optimization problems, demonstrating how mathematical tools can help find solutions that optimize certain criteria under specific constraints.

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Key Takeaways for Optimization Problems

Always clearly define variables and constraints.
Express the target quantity as a function of one variable.
Use derivatives to locate critical points.
Confirm the nature of critical points with the second derivative test.
Consider domain restrictions when interpreting solutions.

By mastering these steps, you can efficiently tackle a wide range of problems involving minimum and maximum values in various fields.

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Meta Description: Discover how to find two numbers with a difference of 42 that produce the minimum product. Learn step-by-step algebraic and calculus methods to solve this optimization problem effectively.

Keywords: find two numbers, minimum product, fixed difference, algebra, calculus, optimization, mathematical problem-solving, minimize product with difference 42