Problem 334 - Calculus Volume 3

Question

For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these. $$f(x, y)=x^{2}+x y+y^{2}-x-y+1$$

Step-by-Step Solution

Answer

The critical point \((\frac{2}{3}, -\frac{1}{3})\) is a local minimum.

Step 1: Find the First Derivatives

To find the critical points, first compute the partial derivatives of the function. The partial derivative with respect to \(x\) is \( f_x = \frac{\partial}{\partial x}(x^2 + xy + y^2 - x - y + 1) = 2x + y - 1 \). The partial derivative with respect to \(y\) is \( f_y = \frac{\partial}{\partial y}(x^2 + xy + y^2 - x - y + 1) = x + 2y - 1 \).

Step 2: Solve for Critical Points

Set the partial derivatives equal to zero to solve for the critical points:1. \(2x + y - 1 = 0\)2. \(x + 2y - 1 = 0\)Substituting the first equation into the second, solve for \(x\) and \(y\). From equation (1), \(y = 1 - 2x\). Substitute into equation (2):\(x + 2(1 - 2x) - 1 = 0\), \(-3x + 2 = 0\),\(x = \frac{2}{3}\),and substituting back into \(y = 1 - 2x\),\(y = 1 - \frac{4}{3} = -\frac{1}{3}\). Hence, the critical point is \( (\frac{2}{3}, -\frac{1}{3}) \).

Step 3: Find the Second Derivatives

Find the second partial derivatives needed for the second derivative test:\( f_{xx} = \frac{\partial^2}{\partial x^2}(x^2 + xy + y^2 - x - y + 1) = 2 \),\( f_{yy} = \frac{\partial^2}{\partial y^2}(x^2 + xy + y^2 - x - y + 1) = 2 \),\( f_{xy} = \frac{\partial^2}{\partial x \partial y}(x^2 + xy + y^2 - x - y + 1) = 1 \).

Step 4: Compute the Discriminant

The discriminant \(D\) is given by the formula:\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]Substitute the values:\[ D = (2)(2) - (1)^2 = 4 - 1 = 3 \]

Step 5: Apply the Second Derivative Test

Since \(D > 0\) and \(f_{xx} > 0\), the critical point \(\left( \frac{2}{3}, -\frac{1}{3} \right)\) is a local minimum of the function \(f(x, y)\).

Key Concepts

Critical Points

Critical points are where a function's rate of change slows down or stops, making them crucial for finding local maxima, minima, or saddle points. To identify these points, we find where the first partial derivatives of the function equal zero.
For our function, the partial derivatives are:

• \( f_x = 2x + y - 1 \)
• \( f_y = x + 2y - 1 \)

To locate the critical points, solve for when both \( f_x \) and \( f_y \) are zero. Solving these equations allows us to find that the critical point for this function is \( \left( \frac{2}{3}, -\frac{1}{3} \right) \).

Partial Derivatives

Partial derivatives are used to explore the slope or rate of change in multivariable functions, focusing on change with respect to one variable while holding others constant.
For the function \( f(x, y) = x^2 + xy + y^2 - x - y + 1 \), we differentiate partially with respect to \(x\) and \(y\). This determines how the function changes around a point in each direction.
Here, the first partial derivatives,

• \( f_x = 2x + y - 1 \)
• \( f_y = x + 2y - 1 \)

reveal potential critical points when set to zero. These derivatives provide the equations to solve for possible critical points such as \( \left( \frac{2}{3}, -\frac{1}{3} \right) \).

Discriminant

The discriminant aids in classifying critical points using the second derivative test. It's calculated using the second partial derivatives:

• \( f_{xx} = 2 \)
• \( f_{yy} = 2 \)
• \( f_{xy} = 1 \)

Using these, the discriminant \( D \) is given by:\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]Substituting the values,\[ D = (2)(2) - (1)^2 = 3 \]A positive \(D > 0\) indicates that the critical point could be a local minimum or maximum, refined further by checking \( f_{xx} \).

Local Minimum

A local minimum occurs at a critical point if the function decreases in all directions from that point and the second derivative test confirms it.
Using our function, after calculating the discriminant (D = 3) and ensuring \( D > 0 \), we check the sign of \( f_{xx} \). Since \( f_{xx} = 2 > 0 \), it indicates that the critical point \( \left( \frac{2}{3}, -\frac{1}{3} \right) \) is indeed a local minimum.
This means that around the point \( \left( \frac{2}{3}, -\frac{1}{3} \right) \), the function \( f(x, y) \) curves upwards, confirming a local low point.