Problem 324
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. $$ f(x, y)=x^{3}+y^{3}-300 x-75 y-3 $$
Step-by-Step Solution
Verified Answer
Critical points: (10,5) - local minimum, (10,-5) & (-10,5) - saddle points, (-10,-5) - local maximum.
1Step 1: Find First Partial Derivatives
To find critical points, first compute the first partial derivatives of the function. The partial derivatives are \( f_x = \frac{\partial}{\partial x}(x^3 + y^3 - 300x - 75y - 3) \) and \( f_y = \frac{\partial}{\partial y}(x^3 + y^3 - 300x - 75y - 3) \). Calculating these, \( f_x = 3x^2 - 300 \) and \( f_y = 3y^2 - 75 \).
2Step 2: Solve for Critical Points
Find the critical points by setting the first partial derivatives equal to zero. For \( f_x = 0 \): \( 3x^2 - 300 = 0 \). Solving gives \( x^2 = 100 \) or \( x = \pm 10 \). For \( f_y = 0 \): \( 3y^2 - 75 = 0 \). Solving gives \( y^2 = 25 \) or \( y = \pm 5 \). The critical points are \((x, y) = (10, 5), (10, -5), (-10, 5), (-10, -5)\).
3Step 3: Compute Second Partial Derivatives
Next, compute the second partial derivatives: \( f_{xx} = \frac{\partial^2}{\partial x^2}(3x^2 - 300) = 6x \), \( f_{yy} = \frac{\partial^2}{\partial y^2}(3y^2 - 75) = 6y \), and \( f_{xy} = \frac{\partial^2}{\partial x \partial y}(0) = 0 \).
4Step 4: Apply Second Derivative Test
Use the second derivative test: compute \( D = f_{xx}f_{yy} - (f_{xy})^2 \). For each critical point:- For \((10, 5)\): \( D = (6 \times 10)(6 \times 5) - 0^2 = 360 > 0\), \( f_{xx} = 60 > 0 \). So it's a local minimum.- For \((10, -5)\): \( D = (6 \times 10)(6 \times -5) - 0^2 = -360 < 0\). So it's a saddle point.- For \((-10, 5)\): \( D = (6 \times -10)(6 \times 5) - 0^2 = -360 < 0\). So it's a saddle point.- For \((-10, -5)\): \( D = (6 \times -10)(6 \times -5) - 0^2 = 360 > 0\), \( f_{xx} = -60 < 0 \). So it's a local maximum.
Key Concepts
Critical PointsPartial DerivativesSaddle PointLocal MaximumLocal Minimum
Critical Points
Critical points are where the first partial derivatives of a function equal zero. These points are important because they are potential spots for local maxima, minima, or saddle points.
For a function of two variables, such as \( f(x, y) \), critical points occur when both \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 0 \).
In our example \( f(x, y) = x^3 + y^3 - 300x - 75y - 3 \), the critical points were found by solving \( 3x^2 - 300 = 0 \) and \( 3y^2 - 75 = 0 \). This gave the critical points at \((x, y) = (10, 5), (10, -5), (-10, 5), (-10, -5)\).
These points need further investigation to determine their nature, which leads us to partial derivatives.
For a function of two variables, such as \( f(x, y) \), critical points occur when both \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 0 \).
In our example \( f(x, y) = x^3 + y^3 - 300x - 75y - 3 \), the critical points were found by solving \( 3x^2 - 300 = 0 \) and \( 3y^2 - 75 = 0 \). This gave the critical points at \((x, y) = (10, 5), (10, -5), (-10, 5), (-10, -5)\).
These points need further investigation to determine their nature, which leads us to partial derivatives.
Partial Derivatives
Partial derivatives measure the function's rate of change with respect to one variable while keeping other variables constant.
The first partial derivatives \( f_x \) and \( f_y \) are used to find critical points by setting them equal to zero.
The second partial derivatives \( f_{xx}, f_{yy}, \) and \( f_{xy} \), however, help us test these points further with the second derivative test.
In our case, we calculated \( f_x = 3x^2 - 300 \) and \( f_y = 3y^2 - 75 \) to find the critical points, then \( f_{xx} = 6x \), \( f_{yy} = 6y \), and \( f_{xy} = 0 \) for the second derivative test analysis.
Using these, we assess each critical point's nature with the second derivative test.
The first partial derivatives \( f_x \) and \( f_y \) are used to find critical points by setting them equal to zero.
The second partial derivatives \( f_{xx}, f_{yy}, \) and \( f_{xy} \), however, help us test these points further with the second derivative test.
In our case, we calculated \( f_x = 3x^2 - 300 \) and \( f_y = 3y^2 - 75 \) to find the critical points, then \( f_{xx} = 6x \), \( f_{yy} = 6y \), and \( f_{xy} = 0 \) for the second derivative test analysis.
Using these, we assess each critical point's nature with the second derivative test.
Saddle Point
A saddle point is an interesting type of critical point where the function behaves like a saddle.
At a saddle point, the function does not have a local maximum or minimum, but rather it curves up in one direction and down in another.
In terms of the second derivative test, a saddle point occurs when \( D = f_{xx}f_{yy} - (f_{xy})^2 < 0 \).
From our example, both points \((10, -5)\) and \((-10, 5)\) turned out to be saddle points as \( D < 0 \) for them.
This indicates the function curves in opposing directions at these points, resembling the shape of a saddle.
At a saddle point, the function does not have a local maximum or minimum, but rather it curves up in one direction and down in another.
In terms of the second derivative test, a saddle point occurs when \( D = f_{xx}f_{yy} - (f_{xy})^2 < 0 \).
From our example, both points \((10, -5)\) and \((-10, 5)\) turned out to be saddle points as \( D < 0 \) for them.
This indicates the function curves in opposing directions at these points, resembling the shape of a saddle.
Local Maximum
A local maximum is a critical point where the function reaches a peak compared to nearby points.
Think of climbing a hill and reaching a point where every directional step leads downward. That's a local maximum.
For a local maximum via the second derivative test, we find \( D > 0 \) and \( f_{xx} < 0 \).
In our problem, \((-10, -5)\) was determined to be a local maximum because \( D > 0 \) and \( f_{xx} = -60 < 0 \).
This indicates the function has achieved a peak at this location relative to its surroundings.
Think of climbing a hill and reaching a point where every directional step leads downward. That's a local maximum.
For a local maximum via the second derivative test, we find \( D > 0 \) and \( f_{xx} < 0 \).
In our problem, \((-10, -5)\) was determined to be a local maximum because \( D > 0 \) and \( f_{xx} = -60 < 0 \).
This indicates the function has achieved a peak at this location relative to its surroundings.
Local Minimum
A local minimum, on the other hand, is where the function reaches a valley relative to its nearby values.
Imagine descending into a trough where any movement would result in climbing up again. That spot is a local minimum.
The second derivative test reveals a local minimum when \( D > 0 \) and \( f_{xx} > 0 \).
Our function has a local minimum at the critical point \((10, 5)\), because here \( D > 0 \) and \( f_{xx} = 60 > 0 \).
This indicates a minimum point, meaning going away from this spot in any direction increases the function's value.
Imagine descending into a trough where any movement would result in climbing up again. That spot is a local minimum.
The second derivative test reveals a local minimum when \( D > 0 \) and \( f_{xx} > 0 \).
Our function has a local minimum at the critical point \((10, 5)\), because here \( D > 0 \) and \( f_{xx} = 60 > 0 \).
This indicates a minimum point, meaning going away from this spot in any direction increases the function's value.
Other exercises in this chapter
Problem 322
For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, sad
View solution Problem 323
For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, sad
View solution Problem 325
For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, sad
View solution Problem 326
For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, sad
View solution