Problem 46 - Algebra for College Students

Question

Find two numbers whose sum is 30 , such that the sum of the square of one number plus ten times the other number is a minimum.

Step-by-Step Solution

Answer

The numbers are 5 and 25.

Step 1: Define the Variables

Let the two numbers be \( x \) and \( y \). We know that their sum is 30, so we can write the equation \( x + y = 30 \).

Step 2: Express One Variable in Terms of the Other

From the equation \( x + y = 30 \), we can express \( y \) in terms of \( x \): \( y = 30 - x \).

Step 3: Define the Function to Minimize

We want to find a minimum for the expression given, which is the sum of the square of one number plus ten times the other. This can be written as the function \( f(x) = x^2 + 10y \). Substitute \( y = 30 - x \) into this function, yielding \( f(x) = x^2 + 10(30-x) \).

Step 4: Simplify the Function

Substitute \( y = 30 - x \) into the function \( f(x) = x^2 + 10y \). This gives \( f(x) = x^2 + 10(30-x) = x^2 + 300 - 10x \). Simplify to \( f(x) = x^2 - 10x + 300 \).

Step 5: Find the Critical Points

To find the minimum, we take the derivative of \( f(x) \), which is \( f'(x) = 2x - 10 \). Set the derivative to zero to find critical points: \( 2x - 10 = 0 \). Solving this gives \( x = 5 \).

Step 6: Verify the Minimum Value

To confirm it's a minimum, check the second derivative \( f''(x) = 2 \). Since \( f''(x) > 0 \), the function is concave up, and \( x = 5 \) is a point of minimum. Substitute back to find \( y = 30 - x = 25 \).

Step 7: Conclude the Results

The two numbers that satisfy the conditions are \( x = 5 \) and \( y = 25 \). The minimum value of the function is \( f(5) = 5^2 - 10\times5 + 300 = 275 \).

Key Concepts

Critical Points

In algebra, finding critical points is essential to solving optimization problems. Critical points are values where the derivative of a function is zero or undefined. They often indicate locations where a function's maximum or minimum values occur.
When dealing with algebraic expressions, you typically find critical points by following these steps:

• Take the derivative of the given function.
• Set the derivative equal to zero.
• Solve this equation to find the critical points.

For instance, if we're working with the function defined in the original problem, the critical points were determined by setting the derivative of the function to zero and solving for the variable, which gives a point where the function can potentially reach a minimum or maximum value.

Derivative Test

The derivative test is a handy tool in calculus for identifying local maxima and minima. By using the first and possibly the second derivative of a function, we can determine the nature of critical points.
The first derivative test involves checking the sign of the derivative before and after each critical point:

• If the sign changes from positive to negative, there's a local maximum.
• If it changes from negative to positive, there's a local minimum.
• If there's no change, it's possibly a point of inflection.

The second derivative test is often used to confirm these findings:

• Positive second derivative (\( f''(x) > 0 \)) indicates a minimum.
• Negative second derivative (\( f''(x) < 0 \)) indicates a maximum.

In the original exercise, the second derivative was positive, thus confirming a local minimum.

Function Minimization

Function minimization is about finding input values for which a function takes on its smallest value. This process is closely tied to the concept of critical points and derivative testing.
For the function \( f(x) = x^2 - 10x + 300 \) from the exercise, finding the minimum involved finding its critical points and testing them. Minimizing functions can help solve practical problems, like cost reduction in business or construction optimizations.
After determining the critical point where the function reaches its minimum value, you would also substitute back to calculate the corresponding dependent variable. In our case, the function minimized at \( x = 5 \) after testing the second derivative to check its nature.

Quadratic Functions

Quadratic functions are polynomial functions of degree two, generally of the form \( f(x) = ax^2 + bx + c \). These functions can model various real-life scenarios such as projectile motion and area problems.
They graph as parabolas, which are symmetric curves that either open upwards or downwards. The vertex of a parabola represents the function's maximum or minimum value, depending on its orientation.
In algebra, quadratic functions are often solved through factorizations, completing the square, or using the quadratic formula. In the case examined, the quadratic expression simplified within the optimization problem helped us identify the minimum sum of squares and linear terms, playing a crucial role in reaching the solution of the problem.