Answer:
Explanation:
To find the critical points (where the partial derivatives are zero or undefined) of the function f(x, y), we need to set the partial derivatives equal to zero and solve the resulting system of equations. Then, we can use the second partial derivative test to determine whether each critical point is a maximum, minimum, or saddle point. Let's analyze each case:
a. tₓ = 3x - 6y and tᵧ = 3y - 6x
Setting tₓ = 0 and tᵧ = 0, we have the following system of equations:
3x - 6y = 0 ...(1)
3y - 6x = 0 ...(2)
To solve this system, we can rearrange equation (1) to express x in terms of y:
3x = 6y
x = 2y ...(3)
Substituting equation (3) into equation (2), we can solve for y:
3y - 6(2y) = 0
3y - 12y = 0
-9y = 0
y = 0
Substituting y = 0 into equation (3), we find:
x = 2(0)
x = 0
Therefore, the critical point is (0, 0).
Now, let's use the second partial derivative test to determine the nature of the critical point:
The discriminant D = fₓₓ(x, y)fᵧᵧ(x, y) - (fₓᵧ(x, y))², where fₓₓ is the second partial derivative of f with respect to x, fᵧᵧ is the second partial derivative of f with respect to y, and fₓᵧ is the second partial derivative of f with respect to x and y.
Calculating the second partial derivatives of f(x, y):
fₓₓ = ∂²f/∂x² = 0
fᵧᵧ = ∂²f/∂y² = 0
fₓᵧ = ∂²f/∂x∂y = 0
Plugging these values into the discriminant:
D = 0(0) - (0)² = 0
Since the discriminant D is zero, the second partial derivative test is inconclusive.
Therefore, at the critical point (0, 0), we cannot determine whether it is a maximum, minimum, or saddle point.
b. tₓ
It seems there is a partial derivative missing in your question. Please provide the complete partial derivative equation (tₓ) for part b so that I can assist you further.