Problem 14 - Calculus and its applications

Question

Find the relative maximum and minimum values. $$ f(x, y)=e^{x^{2}-2 x+y^{2}-4 y+2} $$

Step-by-Step Solution

Answer

The function has a relative minimum at (1, 2) with value \( e^{-3} \).

Step 1: Find First Order Partial Derivatives

To find critical points, we begin by finding the first-order partial derivatives of the function with respect to both variables, \(x\) and \(y\). For \(x\): \[ f_x = \frac{\partial}{\partial x} \left( e^{x^2 - 2x + y^2 - 4y + 2} \right) = e^{x^2 - 2x + y^2 - 4y + 2} \cdot (2x - 2) \]For \(y\): \[ f_y = \frac{\partial}{\partial y} \left( e^{x^2 - 2x + y^2 - 4y + 2} \right) = e^{x^2 - 2x + y^2 - 4y + 2} \cdot (2y - 4) \]

Step 2: Set Partial Derivatives to Zero and Solve

Set the partial derivatives equal to zero to find critical points. For \(f_x = 0\): \[ (2x - 2) e^{x^2 - 2x + y^2 - 4y + 2} = 0 \]This gives: \[ 2x - 2 = 0 \Rightarrow x = 1 \]For \(f_y = 0\): \[ (2y - 4) e^{x^2 - 2x + y^2 - 4y + 2} = 0 \]This gives: \[ 2y - 4 = 0 \Rightarrow y = 2 \]Thus, the critical point is \((x, y) = (1, 2)\).

Step 3: Evaluate the Second Derivative Test

The second derivative test helps us determine the nature of the critical point. First, calculate the second-order partial derivatives. \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x} (2x - 2)e^{x^2 - 2x + y^2 - 4y + 2} = (2 + (4x - 4)^2)e^{x^2 - 2x + y^2 - 4y + 2} \)\( f_{yy} = \frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y} (2y - 4)e^{x^2 - 2x + y^2 - 4y + 2} = (2 + (4y - 8)^2)e^{x^2 - 2x + y^2 - 4y + 2} \)\( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial y} (2x - 2)e^{x^2 - 2x + y^2 - 4y + 2} = (4x - 4)(4y - 8)e^{x^2 - 2x + y^2 - 4y + 2} \)

Step 4: Determine the Nature of the Critical Point

Compute the determinant of the Hessian matrix, \(D\), at the critical point \((1, 2)\):\[ D = f_{xx}(1, 2)f_{yy}(1, 2) - [f_{xy}(1, 2)]^2 \]Given that the exponential term simplifies to 1 at the point \((1, 2)\), we focus on the coefficients:\[ D = (2 + 0)(2 + 0) - (0)^2 = 4 \]Since \(D > 0\) and \(f_{xx} > 0\), the function has a relative minimum at \((1, 2)\).

Step 5: Conclusion

After evaluating with the second derivative test, we determine that the function has a relative minimum at the point \((1, 2)\). The relative minimum value can be calculated by plugging \((1, 2)\) back into the original function: \[ f(1, 2) = e^{1 - 2 + 4 - 8 + 2} = e^{-3} \]

Key Concepts

Partial Derivatives

Partial derivatives are fundamental concepts in multivariable calculus, allowing us to understand how a multivariable function changes as we vary one of its variables while keeping others constant. If you can imagine walking across a hilly landscape, two variables, such as your position east-west and north-south, might determine your elevation. Partial derivatives help by measuring the slope of the hill in one direction while ignoring movement in the other direction. This approach helps us examine how a function reacts to a slight change in only one coordinate.

When calculating a partial derivative with respect to a variable, you treat all other variables as constants. For example, to find the partial derivative of a function \( f(x, y) = e^{x^2 - 2x + y^2 - 4y + 2} \) with respect to \( x \), you get:

• \( f_x = \frac{\partial}{\partial x}(e^{x^2 - 2x + y^2 - 4y + 2}) = e^{x^2 - 2x + y^2 - 4y + 2} \cdot (2x - 2) \)

Similarly, the partial derivative with respect to \( y \) gives:

• \( f_y = \frac{\partial}{\partial y}(e^{x^2 - 2x + y^2 - 4y + 2}) = e^{x^2 - 2x + y^2 - 4y + 2} \cdot (2y - 4) \)

By setting these derivatives to zero, we find points where the slope in one direction is flat, hinting at potential extreme points, either a maximum, minimum, or saddle point.

Critical Points

Critical points are where the first-order partial derivatives equal zero, indicating flatness in all directions of interest at that point. Essentially, you're looking for spots on a bumpy surface where the slope levels out—where the hill neither climbs nor falls drastically no matter which way you look.

After obtaining the partial derivatives of the function, the next step is to set these derivatives to zero:

• For \( f_x = 0 \): \( 2x - 2 = 0 \), which simplifies to \( x = 1 \).
• For \( f_y = 0 \): \( 2y - 4 = 0 \), which simplifies to \( y = 2 \).

This process identifies the critical point \( (1, 2) \). At this point, the surface stops climbing or descending in the directions determined by the variables \( x \) and \( y \), and could be a peak, a trough, or a flattening saddle point. Identifying critical points is crucial, as they are likely candidates for local optima.

Second Derivative Test

The second derivative test provides a way to classify critical points we have already discovered. This test decides if a critical point is a local maximum, a local minimum, or a saddle point by examining the concavity or convexity of the surface. The process involves calculating the second-order partial derivatives.For the function \( f(x, y) = e^{x^2 - 2x + y^2 - 4y + 2} \), these second derivatives are:

• \( f_{xx} = (2 + (4x - 4)^2)e^{x^2 - 2x + y^2 - 4y + 2} \)
• \( f_{yy} = (2 + (4y - 8)^2)e^{x^2 - 2x + y^2 - 4y + 2} \)
• \( f_{xy} = (4x - 4)(4y - 8)e^{x^2 - 2x + y^2 - 4y + 2} \)

The determinant of the Hessian matrix, \( D \), is then computed using these second derivatives. At the critical point \( (1, 2) \), simplifies to:

• \( D = (2)(2) - (0)^2 = 4 \)

If \( D > 0 \) and \( f_{xx} > 0 \), the critical point is a local minimum. If \( D > 0 \) and \( f_{xx} < 0 \), it is a local maximum. If \( D < 0 \), the critical point is a saddle point. With \( D = 4 \) and \( f_{xx} > 0 \), our critical point \( (1, 2) \) is confirmed to be a local minimum. Evaluating \( f(1, 2) \) gives the local minimum value \( e^{-3} \). This method is vital for distinguishing the nature of a point on complex surfaces.