Problem 44 - Intermediate Algebra

Question

Use quadratic functions. Find two numbers whose sum is 50 and whose product is a maximum.

Step-by-Step Solution

Answer

The two numbers are 25 and 25.

Step 1: Define the Problem Using a Quadratic Equation

We need to find two numbers, say \(x\) and \(y\), such that their sum is 50 and their product is maximized. First, we set up the equation for the sum: \(x + y = 50\). We can solve for \(y\) in terms of \(x\), resulting in \(y = 50 - x\).

Step 2: Express the Product as a Function

To find the maximum product, express the product \(P\) of the numbers in terms of \(x\): \[P = x imes y = x (50 - x) = 50x - x^2\]. This is a quadratic function in the form \(P(x) = -x^2 + 50x\).

Step 3: Determine the Vertex of the Quadratic Function

The quadratic function \(P(x) = -x^2 + 50x\) opens downward (since the leading coefficient is negative), so its maximum value occurs at the vertex. For a quadratic \(ax^2 + bx + c\), the x-coordinate of the vertex is given by \(-\frac{b}{2a}\). Substitute \(a = -1\) and \(b = 50\):\[ x = -\frac{50}{2 \times -1} = 25\].

Step 4: Find the Pair of Numbers

Substitute \(x = 25\) back into the equation for \(y\): \[ y = 50 - x = 50 - 25 = 25\].Thus, the two numbers are both equal to 25.

Step 5: Verify the Solution

Verify that the sum of the numbers is 50: \(25 + 25 = 50\), which is correct. The product is \(25 \times 25 = 625\), which is the maximum product possible under these conditions.

Key Concepts

Maximum Product

The concept of finding a maximum product involves optimizing a scenario where you multiply two numbers to get the largest possible result. Imagine you are given a constraint, like the sum of the numbers must be 50. You then express their product as a quadratic function. By using algebra, particularly the properties of quadratic equations, you can discover where this product reaches its maximum. The task is to craft an equation or a quadratic function that represents your product, then analyze it to identify its peak value. This way, you ensure you are getting the maximum multiplication result that meets the given criteria.

Vertex of Quadratic Function

The vertex of a quadratic function is a pivotal point that determines the function's maximum or minimum value. When dealing with the standard form of a quadratic equation, \(ax^2 + bx + c\), you can find the x-value of the vertex using the formula \(-\frac{b}{2a}\). It's important to know the direction in which your parabola opens: upward for positive \(a\) and downward for negative \(a\). In terms of finding a maximum product, our focus is on parabolas that open downward because these contain the maximum points at their vertices.

Once you find the x-coordinate of the vertex, you can substitute it back into the equation to find the paired number (y-coordinate) if needed. This step is crucial in solving real-world problems like maximizing product under given constraints.

Solving Quadratic Equations

Understanding how to solve quadratic equations is essential for working with scenarios involving maximum or minimum values. You often manipulate these equations to model real-world problems, such as finding maximum product or minimizing costs. The standard approach to solving quadratic equations involves:

• Setting up the quadratic equation in the form \(ax^2 + bx + c = 0\).
• Using methods like factoring, completing the square, or applying the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots or solutions.

However, when focusing on the vertex for maximization/minimization purposes, you're more interested in identifying the turning point instead of roots. This understanding enlightens how the quadratic function behaves, allowing you to make strategic decisions based on its properties.