Problem 11 - A Graphical Approach to College Algebra

Question

Solve each problem. Sum of Two Numbers Suppose that \(x\) represents one of two positive numbers whose sum is 30 (a) Represent the other of the two numbers in terms of \(x .\) (b) What are the restrictions on \(x ?\) (c) Determine a function \(P\) that represents the product of the two numbers. (d) Determine analytically and support graphically the two such numbers whose product is a maximum. What is this maximum product?

Step-by-Step Solution

Answer

The other number is \(30 - x\), \(0 < x < 30\), the product function is \(P(x) = 30x - x^2\), and the maximum product of 225 occurs when both numbers are 15.

Step 1: Represent the Other Number

Since the sum of the two numbers is given as 30, we can express the second number in terms of \(x\). Let's call the second number \(y\). Therefore, we have the equation: \[ x + y = 30 \]From this equation, solve for \(y\):\[ y = 30 - x \] Thus, the other number expressed in terms of \(x\) is \(y = 30 - x\).

Step 2: Determine Restrictions on x

The numbers are both positive, so both \(x\) and \(y\) must be greater than 0:1. \(x > 0\)2. \(30 - x > 0\)From \(30 - x > 0\), we get \(x < 30\).Thus, the restrictions on \(x\) are: \[ 0 < x < 30 \]

Step 3: Formulate the Product Function

The product \(P\) of the two numbers \(x\) and \(y\) is given by:\[ P = x imes y \]Substitute \(y = 30 - x\) from Step 1 into the equation:\[ P = x(30 - x) \]This simplifies to:\[ P(x) = 30x - x^2 \]Therefore, the function representing the product is \(P(x) = 30x - x^2\).

Step 4: Find Maximum Product Analytically

To find the maximum product, we need to find the critical points of the function \(P(x) = 30x - x^2\). First, find the derivative:\[ P'(x) = 30 - 2x \]Set \(P'(x)\) to zero to find the critical points:\[ 30 - 2x = 0 \]Solve for \(x\):\[ 2x = 30 \]\[ x = 15 \]The critical point within the domain \(0 < x < 30\) is \(x = 15\). Use the second derivative test or verify the result by comparing \(P(x)\) values at the boundary of the domain. The second derivative is:\[ P''(x) = -2 \]Since \(P''(x) < 0\), \(x = 15\) is a local maximum point.

Step 5: Graphical Confirmation

Graph the function \(P(x) = 30x - x^2\) within the domain \(0 < x < 30\). The graph is a downward-opening parabola with a vertex at the maximum point \(x = 15\), confirming the maximum product occurs at \(x = 15\), \(y = 30 - 15 = 15\).The maximum product is:\[ P(15) = 15 \times 15 = 225 \]

Key Concepts

Sum of Two Numbers

When given the task to deal with the sum of two numbers, it often means we need to express one number in terms of another, especially if their total is known. Here, we have two numbers where the sum is provided as 30. Let's denote the first number as \(x\). Thus, the expression for the sum of the two numbers can be written as:

• \(x + y = 30\)

From this equation, if we need to express the second number, \(y\), in terms of the first number, \(x\), we rearrange the equation to get:

• \(y = 30 - x\)

This allows us to easily work with the second number by substituting its equivalent expression whenever needed.

Product Function

To find the product of two numbers where one is already expressed in terms of the other, we form what's called a product function. The product \(P\) of the two numbers can be represented as the multiplication of one number by the other.

• \(P = x \times y\)

Substituting the expression for \(y\) from before, the product becomes:

• \(P = x(30 - x)\)

Simplifying, it results in the function:

• \(P(x) = 30x - x^2\)

This quadratic equation will enable us to find important values such as the maximum product of the two numbers.

Maximum Product

To find the maximum product, we need to explore the highest point of the quadratic function we've derived, using calculus techniques. The function \(P(x) = 30x - x^2\) describes a parabola that opens downwards, which means it has a peak or maximum point. To find this peak, we calculate the derivative of \(P(x)\):

• \(P'(x) = 30 - 2x\)

Setting the derivative equal to zero helps find the critical points:

• \(30 - 2x = 0\)
• \(x = 15\)

Since the second derivative \(P''(x) = -2\) is negative, it confirms that \(x = 15\) is indeed a local maximum. Thus, at \(x = 15\) and \(y = 15\), the maximum product is \(15 \times 15 = 225\).

Restrictions on Variables

In problems like these, it's essential to define restrictions on the variables involved. The problem specifies that both numbers must be positive. Thus, this condition provides two inequalities:

• \(x > 0\)
• \(y > 0\)

With \(y = 30 - x\), the condition for \(y\) gives us:

• \(30 - x > 0\) which translates to \(x < 30\)

Thus, the restrictions on \(x\) are that it must lie between 0 and 30, strictly:

• \(0 < x < 30\)

These constraints help us understand the domain within which our function and subsequent calculations remain valid.