Problem 4 - Essential Calculus Early Transcendentals

Question

The sum of two positive numbers is \(16 .\) What is the smallest possible value of the sum of their squares?

Step-by-Step Solution

Answer

The smallest possible value of the sum of their squares is 128.

Step 1: Define the Variables

Let's say the two positive numbers are \(x\) and \(y\). According to the problem, the sum of these two numbers is \(x + y = 16\).

Step 2: Express One Variable in Terms of the Other

Since \(x + y = 16\), we can express \(y\) in terms of \(x\) as follows: \(y = 16 - x\).

Step 3: Write the Sum of the Squares Function

We want to minimize the sum of the squares of these two numbers, which is \(x^2 + y^2\). Replacing \(y\) with \(16 - x\), the function becomes \(x^2 + (16 - x)^2\).

Step 4: Simplify the Function

Expand the expression in the function: \[x^2 + (16 - x)^2 = x^2 + (256 - 32x + x^2) = 2x^2 - 32x + 256.\]

Step 5: Find the Derivative

To find the minimum value of the function, compute the derivative of \(f(x) = 2x^2 - 32x + 256\). The derivative is \[f'(x) = 4x - 32.\]

Step 6: Find the Critical Points

Set the derivative equal to zero to find the critical points: \[4x - 32 = 0.\] Solving for \(x\), we find \(x = 8\).

Step 7: Verify the Minimum Value

Since the function is a quadratic with a positive leading coefficient, it opens upwards, meaning the critical point found at \(x = 8\) is a minimum. Substituting back \(x = 8\) into \(y = 16 - x\), we get \(y = 8\).

Step 8: Calculate the Minimum Sum of Squares

Substitute \(x = 8\) and \(y = 8\) into the sum of squares: \[x^2 + y^2 = 8^2 + 8^2 = 64 + 64 = 128.\]

Key Concepts

Quadratic Functions

Quadratic functions are a fundamental part of calculus and algebra, represented in the form of a polynomial equation:

• The general format is \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\).
• These functions graph as parabolas, which are U-shaped curves that open upwards if \(a > 0\) and downwards if \(a < 0\).
• The vertex of the parabola represents the highest or lowest point, known as the maximum or minimum.

In optimization problems, like finding the minimum sum of two squares, quadratic functions allow us to determine these extreme values, using their specific structure and properties. In this exercise, the sum of the squares, \(x^2 + (16 - x)^2\), simplifies to \(2x^2 - 32x + 256\), a quadratic equation that can be optimized.

Derivative

The derivative of a function provides valuable information about its rate of change and allows us to find key points such as maximums and minimums.

• To derive a function, we calculate how it changes as its input changes.
• For a function \(f(x) = 2x^2 - 32x + 256\), the derivative is found using the power rule and turns out to be \(f'(x) = 4x - 32\).
• This derivative equation tells us how the value of the quadratic function changes with respect to \(x\).

By setting the derivative to zero, \(f'(x) = 0\), we identify where the slope of the tangent to the function equals zero. These points are known as critical points, essential in finding optimization solutions.

Critical Points

Critical points play a vital role in finding the optimal values in calculus, particularly when dealing with quadratic functions.

• A critical point occurs where the derivative of a function equals zero, \(f'(x) = 0\), indicating a potential maximum or minimum point on the graph.
• In the exercise, the derivative \(f'(x) = 4x - 32\) is set to zero, yielding \(x = 8\) as a critical point.
• We determine whether this point is a minimum or maximum by examining the general shape of the quadratic function.
• Since it opens upwards (the coefficient of \(x^2\) is positive), \(x = 8\) represents a minimum.

By substituting this back into the original expression involving \(y = 16 - x\), we determine the other value, \(y = 8\), and confirm that these are the dimensions for which the sum of the squares is minimized, amounting to \(128\). This approach is key to solving optimization problems efficiently.