Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
f(x, y) = x^2 +xy +y^2 +2y
This should get you started.
https://www.wolframalpha.com/input/?i=x%5E2+%2Bxy+%2By%5E2+%2B2y
To find the local maximum and minimum values and saddle points of the function f(x, y) = x^2 + xy + y^2 + 2y, we need to take the partial derivatives with respect to x and y and set them equal to zero.
1. Find the partial derivative with respect to x:
∂f(x, y)/∂x = 2x + y
2. Find the partial derivative with respect to y:
∂f(x, y)/∂y = x + 2y + 2
Now, set both partial derivatives equal to zero and solve for x and y.
1. Setting ∂f(x, y)/∂x = 0:
2x + y = 0
y = -2x
2. Setting ∂f(x, y)/∂y = 0:
x + 2y + 2 = 0
x + 2(-2x) + 2 = 0
6x = -2
x = -1/3
Substitute the value of x into the equation for y:
y = -2(-1/3)
y = 2/3
So, we found a critical point at (-1/3, 2/3).
Now, we need to analyze the nature of this critical point to determine if it is a local maximum, minimum, or saddle point. For this, we use the second partial derivatives test.
1. Find the second partial derivative with respect to x:
∂^2f(x, y)/∂x^2 = 2
2. Find the second partial derivative with respect to y:
∂^2f(x, y)/∂y^2 = 2
3. Find the mixed partial derivative:
∂^2f(x, y)/∂x∂y = 1
Now, use these second partial derivatives to evaluate the discriminant:
D = (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2
D = (2)(2) - (1)^2
D = 3
If D > 0 and (∂^2f/∂x^2) > 0, then the critical point is a local minimum.
If D > 0 and (∂^2f/∂x^2) < 0, then the critical point is a local maximum.
If D < 0, then the critical point is a saddle point.
In this case, since D = 3 and (∂^2f/∂x^2) = 2 (which is greater than zero), we can conclude that the critical point (-1/3, 2/3) is a local minimum.
Therefore, the local minimum value is f(-1/3, 2/3) = (-1/3)^2 + (-1/3)(2/3) + (2/3)^2 + 2(2/3) = 2/3.
As for the graphing aspect of the question, if you have three-dimensional graphing software, you can plot the function f(x, y) = x^2 + xy + y^2 + 2y in the x-y plane to visualize the function and observe the critical point at (-1/3, 2/3). The graph will reveal the shape of the surface and the behavior of the function.
To find the local maximum and minimum values and saddle points of the function f(x, y) = x^2 + xy + y^2 + 2y, we need to find the critical points and evaluate the second-order partial derivatives. Let's go step-by-step:
Step 1: Find the gradient of f(x, y)
The gradient of f(x, y) is given by ∇f(x, y) = [∂f/∂x, ∂f/∂y].
Taking partial derivatives with respect to x and y, we have:
∂f/∂x = 2x + y
∂f/∂y = x + 2y + 2
Step 2: Find the critical points
The critical points occur where the gradient is equal to zero.
Setting ∂f/∂x = 0 and ∂f/∂y = 0, we have the system of equations:
2x + y = 0
x + 2y + 2 = 0
Solving this system of equations, we find the critical point:
x = -2/3, y = 4/3
Step 3: Evaluate the second-order partial derivatives
Now, we need to compute the second-order partial derivatives for the Hessian matrix.
∂²f/∂x² = 2
∂²f/∂y² = 2
And,
∂²f/∂x∂y = ∂²f/∂y∂x = 1
Step 4: Determine the nature of the critical point
Using the Hessian matrix, we can classify the critical point as either local maximum, minimum, or saddle point.
The determinant of the Hessian matrix is:
D = (∂²f/∂x²)(∂²f/∂y²) - (∂²f/∂x∂y)(∂²f/∂y∂x) = (2)(2) - (1)(1) = 3
The trace of the Hessian matrix is:
T = ∂²f/∂x² + ∂²f/∂y² = 2 + 2 = 4
The discriminant of the Hessian matrix is:
D = T^2 - 4D = 4^2 - 4(3) = 16 - 12 = 4
Using the discriminant, we can determine the nature of the critical point:
If D > 0 and T > 0, then the critical point is a local minimum.
If D > 0 and T < 0, then the critical point is a local maximum.
If D < 0, then the critical point is a saddle point.
For our function, D = 4 > 0 and T = 4 > 0, so the critical point (-2/3, 4/3) is a local minimum.
Therefore, the local minimum is (-2/3, 4/3), and there are no local maximum or saddle points.
Please note that we can't provide a graph as we're text-based AI, but you can use three-dimensional graphing software to visualize the function.