Problem 12 - APEX Calculus

Question

Find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, or saddle point. $$ f(x, y)=x^{4}-2 x^{2}+y^{3}-27 y-15 $$

Step-by-Step Solution

Answer

Critical points: (0,3) saddle, (0,-3) max, (1,3) min, (1,-3) saddle, (-1,3) min, (-1,-3) saddle.

Step 1: Find the First Partial Derivatives

To find the critical points of the function, we first need to compute the first partial derivatives of the function with respect to both variable \(x\) and \(y\). For the function \(f(x, y) = x^4 - 2x^2 + y^3 - 27y - 15\), we calculate:\[ f_x = \frac{\partial}{\partial x}(x^4 - 2x^2 + y^3 - 27y - 15) = 4x^3 - 4x \]\[ f_y = \frac{\partial}{\partial y}(x^4 - 2x^2 + y^3 - 27y - 15) = 3y^2 - 27 \]

Step 2: Solve for Critical Points

Critical points occur where both first partial derivatives are zero. We set \(f_x = 0\) and \(f_y = 0\) to find critical points:For \(f_x = 0\):\[ 4x^3 - 4x = 0 \]Factoring out \(4x\):\[ 4x(x^2 - 1) = 0 \]\[ x(x+1)(x-1) = 0 \]Thus, \(x = 0, x = 1, x = -1\).For \(f_y = 0\):\[ 3y^2 - 27 = 0 \]\[ y^2 = 9 \]\[ y = 3, y = -3 \]The solutions give us potential critical points at \((0, 3), (0, -3), (1, 3), (1, -3), (-1, 3), (-1, -3)\).

Step 3: Find the Second Partial Derivatives

To apply the Second Derivative Test, we need the second partial derivatives:\[ f_{xx} = \frac{\partial^2}{\partial x^2}(4x^3 - 4x) = 12x^2 - 4 \]\[ f_{yy} = \frac{\partial^2}{\partial y^2}(3y^2 - 27) = 6y \]\[ f_{xy} = \frac{\partial^2}{\partial x \partial y}(0) = 0 \]

Step 4: Apply Second Derivative Test

The Second Derivative Test involves the Hessian determinant:\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]Calculate \(D\) and the sign of \(f_{xx}\) for each critical point:1. At \((0, 3)\): \[ f_{xx} = -4, \quad f_{yy} = 18, \quad D = (-4)(18) - 0^2 = -72 \] \(D < 0\), so it's a saddle point.2. At \((0, -3)\): \[ f_{xx} = -4, \quad f_{yy} = -18, \quad D = (-4)(-18) - 0^2 = 72 \] \(D > 0\) and \(f_{xx} < 0\), indicating a local maximum.3. At \((1, 3)\): \[ f_{xx} = 8, \quad f_{yy} = 18, \quad D = (8)(18) - 0^2 = 144 \] \(D > 0\) and \(f_{xx} > 0\), indicating a local minimum.4. At \((1, -3)\): \[ f_{xx} = 8, \quad f_{yy} = -18, \quad D = (8)(-18) - 0^2 = -144 \] \(D < 0\), so it's a saddle point.5. At \((-1, 3)\): \[ f_{xx} = 8, \quad f_{yy} = 18, \quad D = (8)(18) - 0^2 = 144 \] \(D > 0\) and \(f_{xx} > 0\), indicating a local minimum.6. At \((-1, -3)\): \[ f_{xx} = 8, \quad f_{yy} = -18, \quad D = (8)(-18) - 0^2 = -144 \] \(D < 0\), so it's a saddle point.

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in multivariable calculus. They measure how a function changes as one of its input variables changes, keeping others constant. Consider the function \( f(x, y) = x^4 - 2x^2 + y^3 - 27y - 15 \). To find critical points, we begin by calculating its first partial derivatives. The partial derivative with respect to \( x \), denoted as \( f_x \), involves differentiating the function treating \( y \) as a constant. Similarly, for \( y \), we calculate \( f_y \), treating \( x \) as a constant. We found:

• \( f_x = 4x^3 - 4x \)
• \( f_y = 3y^2 - 27 \)

Setting these derivatives to zero helps us identify points where the function may have relative maxima, minima, or saddle points.

Second Derivative Test

The Second Derivative Test is a method used to determine the nature of critical points in multivariable functions. Once we find the points where both first partial derivatives are zero, we use the second derivatives. For the function \( f(x, y) = x^4 - 2x^2 + y^3 - 27y - 15 \), we calculate:

• \( f_{xx} = 12x^2 - 4 \)
• \( f_{yy} = 6y \)
• \( f_{xy} = 0 \)

We apply these in the Hessian determinant to assess the nature of points identified as critical. The sign and magnitude of these second derivatives help us make conclusions about the function's behavior at these points.

Hessian Determinant

The Hessian determinant, denoted as \( D \), is a crucial part of the Second Derivative Test. It's defined as \( D = f_{xx}f_{yy} - (f_{xy})^2 \). For our function, the mixed partial derivative \( f_{xy} \) is zero, simplifying our task. The value of \( D \) at each critical point tells us:

• If \( D > 0 \) and \( f_{xx} > 0 \), there is a local minimum.
• If \( D > 0 \) and \( f_{xx} < 0 \), there is a local maximum.
• If \( D < 0 \), the point is a saddle point.

This analysis helps us describe the landscape of the function around the critical points.

Saddle Point

A saddle point is a type of critical point that is neither a maximum nor a minimum. In simpler terms, it's like a mountain pass: going along one direction might take you downwards, while heading in a perpendicular direction might take you upwards. At a saddle point, the Hessian determinant \( D \) is negative. This indicates that the function curves up in some directions and down in others around that point. For our function, we identified saddle points at coordinates where \( D < 0 \): specifically at \( (0, 3) \), \( (1, -3) \), and \( (-1, -3) \). These points show how the surface of the graph could saddle, shaping local valleys and peaks around them.