Problem 311 - Calculus 3

Question

For the following exercises, find all critical points. $$ f(x, y)=(3 x-2)^{2}+(y-4)^{2} $$

Step-by-Step Solution

Answer

The critical point is \( \left( \frac{2}{3}, 4 \right) \).

Step 1: Identify the Function

The function given is a function of two variables, denoted as \[ f(x, y) = (3x - 2)^2 + (y - 4)^2. \]We aim to find the critical points of this function.

Step 2: Find the Partial Derivatives

Calculate the partial derivatives of the function with respect to both variables, x and y.The partial derivative of \( f \) with respect to \( x \) is given by:\[ \frac{\partial f}{\partial x} = 2(3x - 2) \cdot 3 = 6(3x - 2). \]The partial derivative of \( f \) with respect to \( y \) is:\[ \frac{\partial f}{\partial y} = 2(y - 4) \cdot 1 = 2(y - 4). \]

Step 3: Set Partial Derivatives to Zero

To find the critical points, set each partial derivative equal to zero. This gives us a system of equations:For \( x \):\[ 6(3x - 2) = 0 \]Solve for \( x \):\[ 3x - 2 = 0 \]\[ 3x = 2 \]\[ x = \frac{2}{3}. \]For \( y \):\[ 2(y - 4) = 0 \]Solve for \( y \):\[ y - 4 = 0 \]\[ y = 4. \]

Step 4: Determine the Critical Point

We now have values for both variables: \( x = \frac{2}{3} \) and \( y = 4 \). This means the critical point of the function is:\[ \left( \frac{2}{3}, 4 \right). \]

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, particularly in dealing with functions of multiple variables. They represent the rate of change of the function regarding one of its variables while keeping the other variables constant. Think of it like focusing on how a function behaves with respect to one variable at a time.
For the given function \( f(x, y) = (3x - 2)^2 + (y - 4)^2 \), you need to find the partial derivatives with respect to both \( x \) and \( y \). This means you calculate how changes in \( x \) and \( y \) affect \( f \) separately.

• The partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), gives you \( 6(3x - 2) \). It's obtained by applying the chain rule to the term \((3x - 2)^2\).
• The partial derivative with respect to \( y \), denoted as \( \frac{\partial f}{\partial y} \), yields \( 2(y - 4) \), obtained by simply differentiating \((y - 4)^2\).

By setting these derivatives to zero, you find values of \( x \) and \( y \) that make these rates of change zero, which correspond to critical points.

Functions of Multiple Variables

Functions of multiple variables, like \( f(x, y) = (3x - 2)^2 + (y - 4)^2 \), depend on more than one independent variable. Such functions can be visualized as surfaces in three-dimensional space, where each point \((x, y)\) corresponds to a specific \( f(x, y) \) value.
Understanding how these functions work requires observing how changes in each variable affect the function's output. Here, you're balancing the influence of \( x \) and \( y \) to find constants where the function is at its peak, valley, or saddle point, which are known as critical points.
These critical points are found by intersecting the directions determined by partial derivatives, pointing out where the function stabilizes. For a quadratic function like the one we've got, the process is straightforward, but it becomes more complex with additional variables or higher-degree polynomials.
Grasping this concept is critical when analyzing more sophisticated functions in fields such as economics, engineering, and physics.

System of Equations

A system of equations emerges when solving for critical points, as each partial derivative set to zero gives us an equation. Here, the given system consists of two equations:

• \( 6(3x - 2) = 0 \) leading to solving \( x = \frac{2}{3} \)
• \( 2(y - 4) = 0 \) leading to solving \( y = 4 \)

To solve a system of equations like this, you'll need to determine the values of \( x \) and \( y \) that simultaneously satisfy both equations. This typically involves rearranging each equation to isolate one variable.
In real-world applications, systems of equations may become quite complex, requiring algebraic techniques or computational software to solve. Here, the simplicity of linear equations makes finding \( x \) and \( y \) straightforward, resulting in the critical point \( \left( \frac{2}{3}, 4 \right) \). Understanding how to manage systems of equations is invaluable for tackling more advanced mathematical problems that involve identifying and interpreting critical points of a function.