Problem 35 - Calculus an Applied Approach

Question

Find three positive numbers \(x, y,\) and \(z\) that satisfy the given conditions. The sum is 30 and the sum of the squares is a minimum.

Step-by-Step Solution

Answer

The values of \(x, y, z\) that minimize the sum of the squares while their sum is 30, are \(x=10, y=10, z=10\).

Step 1: Defining the problem

Let's denote our unknowns as \(x, y, z\). According to the problem, \(x+y+z=30\). Additionally, we want to minimize the sum of squares \(F=x^2 + y^2 + z^2\).

Step 2: Removing one variable

We can simplify our problem by expressing one variable via the others using the equation \(x+y+z=30\). We can keep as variables \(x\) and \(y\), and find \(z\) as \(z=30-x-y\). The function to minimize then becomes \(F=x^2 + y^2 + (30 - x - y)^2\).

Step 3: Finding the derivative

Next, let's find the derivative of the function \(F\) with respect to \(x\) and \(y\), and set them to zero in order to find the minimum. This gives us \( \frac{dF}{dx}=2x-2(30 - x - y)\) and \( \frac{dF}{dy}=2y-2(30 - x - y)\). Setting these to zero gives us the equations: \(3x+y=30\) and \(x+3y=30\).

Step 4: Solving the equation

Solving these linear equations, we find \(x = y = 10\).

Step 5: Finding z

Substituting \(x\) and \(y\) into \(x+y+z=30\), we get \(z=30-10-10=10\).

Key Concepts

Calculus

Calculus is a branch of mathematics that focuses on change and motion. It provides us with powerful tools for understanding the dynamic nature of the world around us. One of the core components of calculus used in optimization problems is finding the derivative. Derivatives tell us about the rate of change of a function. They help identify maxima and minima, which are the highest and lowest points, respectively, on a graph.
When solving optimization problems, such as minimizing or maximizing a function, calculus lets us use derivatives to find critical points where the function's rate of change is zero. These are potential points for a function's minimum or maximum, often achieved by setting the derivative of the function to zero. For our problem, this involves taking the derivative of the function that describes the sum of squares and finding the values of our variables that make these derivatives zero.

Optimization Problem

An optimization problem involves finding the best solution from a set of possible choices. These problems are everywhere in mathematics, engineering, economics, and many other fields. They often involve constraints, or conditions, that solutions must satisfy.
In our problem, we must find three positive numbers whose sum is 30, and where the sum of their squares is minimized. This is typical of an optimization problem where you need to balance competing requirements: achieving a specific sum and minimizing another quantity.

• The challenge is to translate the problem into mathematical terms, which here means creating equations based on the conditions provided.
• In the given exercise, the sum of the squares is the function we aim to minimize while adhering to the constraint that their sum is 30.

A key step in solving such a problem is to express one variable in terms of others. This reduces the number of variables and simplifies the problem, making it more manageable.

Sum of Squares

The sum of squares is a concept often used in various fields, including statistics and optimization. In this context, it refers to the sum of the squares of the individual numbers we need to consider, namely, \(x^2 + y^2 + z^2\).
Why minimize the sum of squares? It's an objective function, a function we aim to decrease as much as possible, subject to certain constraints.

• Minimizing the sum of squares often leads to balanced solutions where each element is close to each other in magnitude, as seen in this problem.
• Additionally, it can reduce the disparity between numbers, which is sometimes a desired property in design and statistical variance reduction.

In the exercise, the solution found was \(x = y = z = 10\), which meets the condition that their sum is 30. This illustrates how balancing the elements can lead to the desired minimum, leveraging the symmetry of the problem and the properties of quadratic functions.