Problem 7 - Calculus with Concepts in Calculus

Question

Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ f(x, y)=x^{2} y-2 x y+2 y^{2}-15 y $$

Step-by-Step Solution

Answer

Critical points: \((5,0)\) and \((-3,0)\) are saddle points, \((1,4)\) is a relative minimum.

Step 1: Find the First Partial Derivatives

To find critical points, we first need to calculate the first partial derivatives of the function \(f(x, y)\) with respect to \(x\) and \(y\). Let's find \(f_x\) and \(f_y\):\[f_x = \frac{\partial}{\partial x}(x^2 y - 2xy + 2y^2 - 15y) = 2xy - 2y\]\[f_y = \frac{\partial}{\partial y}(x^2 y - 2xy + 2y^2 - 15y) = x^2 - 2x + 4y - 15\]

Step 2: Set the First Partial Derivatives to Zero

To find the critical points of \(f(x, y)\), we need to set \(f_x = 0\) and \(f_y = 0\).1. For \(f_x = 0\): \[2xy - 2y = 0 \Rightarrow 2y(x - 1) = 0\] This gives us two possibilities: \(y = 0\) or \(x = 1\).2. For \(f_y = 0\): \[x^2 - 2x + 4y - 15 = 0\]

Step 3: Solve the System of Equations

Using the equations from Step 2, find the values of \(x\) and \(y\) that satisfy both conditions:1. If \(y = 0\), substitute into the second equation: \[x^2 - 2x - 15 = 0\] This quadratic equation can be solved by factoring: \[(x - 5)(x + 3) = 0\] So, \(x = 5\) or \(x = -3\). Thus, critical points are \((5, 0)\) and \((-3, 0)\).2. If \(x = 1\), substitute into the second equation: \[1^2 - 2(1) + 4y - 15 = 0\] Simplifying gives: \[4y - 16 = 0 \Rightarrow y = 4\] Therefore, another critical point is \((1, 4)\).

Step 4: Use the Second Partial Derivative Test

To classify the critical points, compute the second partial derivatives of \(f(x, y)\):\[f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2y\]\[f_{yy} = \frac{\partial^2 f}{\partial y^2} = 4\]\[f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 2x - 2\]For each critical point \((x, y)\), compute:\[D = f_{xx}f_{yy} - (f_{xy})^2\]

Step 5: Evaluate the Test for Each Critical Point

Let's evaluate \(D\) and determine the nature of each critical point:For \((5,0)\):\[f_{xx} = 2(0) = 0, \quad f_{yy} = 4, \quad f_{xy} = 2(5) - 2 = 8\]\[D = (0)(4) - (8)^2 = -64 < 0\]Saddle point.For \((-3,0)\):\[f_{xx} = 0, \quad f_{yy} = 4, \quad f_{xy} = 2(-3) - 2 = -8\]\[D = (0)(4) - (-8)^2 = -64 < 0\]Saddle point.For \((1,4)\):\[f_{xx} = 2(4) = 8, \quad f_{yy} = 4, \quad f_{xy} = 2(1) - 2 = 0\]\[D = (8)(4) - (0)^2 = 32 > 0\]Since \(f_{xx} > 0\), this is a relative minimum.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental tool in multivariable calculus, allowing us to understand how a function changes with respect to one variable while keeping other variables constant. For the function given, we have two variables, \(x\) and \(y\). To find critical points, we calculate the first-order partial derivatives, \(f_x\) and \(f_y\). These represent the rate of change of the function \(f(x, y)\) with respect to \(x\) and \(y\), respectively.

To find \(f_x\):

• Differentiating the function with respect to \(x\) while considering \(y\) as a constant gives: \(f_x = 2xy - 2y\).

To find \(f_y\):

• Differentiating the function with respect to \(y\) while treating \(x\) as a constant yields: \(f_y = x^2 - 2x + 4y - 15\).

Finding critical points involves setting these two partial derivatives to zero and solving for \(x\) and \(y\).

Second Partial Derivative Test

The second partial derivative test helps classify the nature of critical points found from the first partial derivatives. After obtaining the second partial derivatives, \(f_{xx}\), \(f_{yy}\), and \(f_{xy}\), the test uses these to calculate the determinant \(D\):

\[D = f_{xx} f_{yy} - (f_{xy})^2\] The result tells us the type of critical point:

• If \(D > 0\) and \(f_{xx} > 0\), then there is a relative minimum.
• If \(D > 0\) and \(f_{xx} < 0\), it indicates a relative maximum.
• If \(D < 0\), the point is a saddle point.
• If \(D = 0\), the test is inconclusive, and further analysis is needed.

For this specific problem, the different values of \(D\) help us determine the nature of each explored critical point.

Saddle Point

A saddle point is a critical point where the function has properties of both minimum and maximum in different directions. In this exercise, for critical points \((5,0)\) and \((-3,0)\), the determinant \(D < 0\). This finding indicates saddle points at both locations.

Saddle points occur because the curvature of the function changes signs in various directions at these points. Visualizing a saddle point, it resembles the shape of a horse's saddle, providing a clear indication of why it bears the name "saddle point." Recognizing these points is crucial in understanding the topology and behavior of a function's graph.

Relative Maximum Value

A relative maximum value of a function at a critical point implies that the function attains its highest value within a local range when compared to its immediate surroundings. For this exercise, however, none of the critical points Satisfied the condition for a relative maximum value using the second partial derivative test.

Generally speaking, when analyzing a function, the second partial derivative test reveals a relative maximum if \(D > 0\) and \(f_{xx} < 0\). This indicates a downward opening curvature at the critical point, exhibiting a peak. This is akin to a hilltop on the graph of the function.

Relative Minimum Value

A relative minimum value is identified when a function attains its lowest value within a small neighborhood of the critical point. In the given exercise, the critical point \((1,4)\) proved to be a relative minimum value.

Here, \(D > 0\) and \(f_{xx} > 0\), indicating an upward opening curvature at that point. This provides a visual cue of a valley on the graph of the function. Understanding relative minima is vital, particularly in optimization problems, as they indicate a local low point which might be of interest in practical applications.