Problem 11 - Calculus Early Transcendentals

Question

Find all critical points of the following functions. $$f(x, y)=(3 x-2)^{2}+(y-4)^{2}$$

Step-by-Step Solution

Answer

Question: Determine the critical point for the given function $$f(x, y) = (3x-2)^{2}+(y-4)^{2}$$. Answer: The critical point of the given function is \((\frac{2}{3}, 4)\).

Step 1: Find the partial derivatives

To find the critical points, first find the partial derivatives of the function f(x, y) with respect to both x and y: $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left[(3x-2)^{2}+(y-4)^{2}\right]$$ $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\left[(3x-2)^{2}+(y-4)^{2}\right]$$

Step 2: Calculate the partial derivatives

Now, calculate the partial derivatives found in step 1: $$\frac{\partial f}{\partial x} = 2(3x-2)(3) = 6(3x-2)$$ $$\frac{\partial f}{\partial y} = 2(y-4)(1) = 2(y-4)$$

Step 3: Set the partial derivatives to zero

Set both partial derivatives equal to zero and solve for x and y respectively: $$6(3x-2) = 0$$ $$2(y-4) = 0$$

Step 4: Solve for x and y

Solve the equations found in step 3 for x and y: $$3x-2 = 0 \Rightarrow x = \frac{2}{3}$$ $$y-4 = 0 \Rightarrow y = 4$$

Step 5: Combine the coordinates

Combine the x and y values found to obtain the critical point: Critical Point: $$(\frac{2}{3}, 4)$$ So, the critical point of the given function f(x, y) is \((\frac{2}{3}, 4)\).

Key Concepts

Partial Derivative

A partial derivative is a derivative taken with respect to one variable while keeping other variables constant. In multivariable functions, partial derivatives are crucial in understanding how the function changes with a change in each individual variable. For the function \( f(x, y) = (3x-2)^2 + (y-4)^2 \), we calculate the partial derivatives by differentiating with respect to each variable separately. Here's how it works step-by-step:

• For \( \frac{\partial f}{\partial x} \), we treat \( y \) as a constant and only differentiate \( (3x-2)^2 \).
• For \( \frac{\partial f}{\partial y} \), we treat \( x \) as a constant and differentiate \( (y-4)^2 \).

Partial derivatives help identify the rate of change of a function when one variable is altered at a time, making them essential for finding critical points where these rates of change are zero. This approach ultimately leads to understanding the optimization of the function.

Multivariable Calculus

Multivariable calculus extends calculus concepts to functions of more than one variable. Unlike single-variable calculus, it involves concepts such as partial derivatives, gradients, and multiple integrals. Functions like \( f(x, y) = (3x-2)^2 + (y-4)^2 \) are examples of multivariable functions, defined over a domain in the plane. Critical points in this context are where the function's rate of change in any direction is zero. This is explored using gradients, which are vector combinations of partial derivatives. Calculations can get complex, but the fundamental principle is calculating derivatives with respect to each variable and understanding their impacts. Key concepts include:

• Gradient Vector: An extension of the derivative, \( abla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) \), which points in the direction of greatest increase of the function.
• Critical Points: Places where \( abla f = \mathbf{0} \) or undifferentiable, potential locations for local maxima, minima, or saddle points.

Multivariable calculus is all about exploring these interactions and their implications for real-world scenarios.

Optimization

Optimization is the process of finding the maximum or minimum value of a function. In multivariable calculus, determining critical points is a crucial step in this process. These points might represent local maxima, minima, or saddle points. To find the critical points of \( f(x, y) = (3x-2)^2 + (y-4)^2 \), we solve for zeros in its partial derivatives. These zeros indicate spots where the surface is flat in all directions, potential candidates for optimal points. Using the calculated partial derivatives:

• The solution to \( \frac{\partial f}{\partial x} = 0 \) yields \( x = \frac{2}{3} \).
• The solution to \( \frac{\partial f}{\partial y} = 0 \) gives \( y = 4 \).
• Together, these form the critical point \((\frac{2}{3}, 4)\).

After finding critical points, the next step in optimization is determining their nature. Often, this involves the second derivative test or evaluating the function's behavior around these points. Optimization techniques are pivotal in fields ranging from economics to engineering, where one seeks to minimize costs or maximize efficiency.