Find the maximum and minimum values of f(x, y) = xy on the ellipse 8\!x^{2} + y^{2} = 9.
To find the maximum and minimum values of the function f(x, y) = xy on the ellipse 8x^2 + y^2 = 9, we can use the method of Lagrange multipliers.
Step 1: Define the Lagrangian function
The Lagrangian function is defined as L(x, y, λ) = xy + λ(8x^2 + y^2 - 9), where λ is the Lagrange multiplier.
Step 2: Find the partial derivatives of the Lagrangian function
∂L/∂x = y + 16λx
∂L/∂y = x + 2λy
∂L/∂λ = 8x^2 + y^2 - 9
Step 3: Set the partial derivatives equal to zero and solve the system of equations
Setting ∂L/∂x = 0, we have: y + 16λx = 0 ...(1)
Setting ∂L/∂y = 0, we have: x + 2λy = 0 ...(2)
Setting ∂L/∂λ = 0, we have: 8x^2 + y^2 - 9 = 0 ...(3)
Step 4: Solve the system of equations
From equation (1), y = -16λx and substitute into equation (2):
x - 32λ^2x = 0
Simplifying, we get: (1 - 32λ^2)x = 0
Since x cannot be zero (otherwise, y will also be zero), we have: 1 - 32λ^2 = 0
Solving for λ, we get: λ = ± 1/4
Substituting the value of λ into equation (1), we get: y = ± 4x
Substituting the value of x and y into equation (3), we have:
8x^2 + (±4x)^2 - 9 = 0
Simplifying, we get: 64x^2 = 9
Taking the square root, we get: x = ± 3/8
For λ = 1/4, we have x = 3/8 and y = 4(3/8) = 3/2
For λ = -1/4, we have x = -3/8 and y = -4(3/8) = -3/2
Step 5: Evaluate the function f(x, y) = xy at the critical points
For λ = 1/4, f(3/8, 3/2) = (3/8)(3/2) = 9/16
For λ = -1/4, f(-3/8, -3/2) = (-3/8)(-3/2) = 9/16
Step 6: Determine the maximum and minimum values
The maximum value of f(x, y) = xy on the ellipse is 9/16, and the minimum value is also 9/16.
Therefore, the maximum and minimum values of f(x, y) = xy on the ellipse 8x^2 + y^2 = 9 are both 9/16.
To find the maximum and minimum values of the function f(x, y) = xy on the given ellipse 8x^2 + y^2 = 9, we can use the method of Lagrange multipliers.
Step 1: Define the function to be optimized
Let F(x,y,λ) = xy - λ(8x^2 + y^2 - 9), where λ is the Lagrange multiplier.
Step 2: Find the partial derivatives
Calculate the partial derivatives of F with respect to x, y, and λ, and set them equal to zero:
∂F/∂x = y - 16λx = 0
∂F/∂y = x - 2λy = 0
∂F/∂λ = 8x^2 + y^2 - 9 = 0
Step 3: Solve the system of equations
Solve the system of equations formed by these three equations. From the first equation, we can find y = 16λx, and substituting this into the second equation gives x - 32λ^2 x = 0. This implies that x = 0 or 1 - 32λ^2 = 0.
Case 1: x = 0
If x = 0, then from the first equation, y = 0. Substituting these values into the third equation, we get y^2 = 9, which implies y = ±3.
So, we have two potential points (0, -3) and (0, 3).
Case 2: 1 - 32λ^2 = 0
Solving 1 - 32λ^2 = 0, we get λ = ±1/4√2. Substituting these values into the first equation, we get y = 4x/√2, and substituting into the third equation, we get 8x^2 + (4x/√2)^2 = 9. Simplifying, we get 8x^2 + 8x^2 = 9, which gives 16x^2 = 9. This implies x = ±3/4√2.
Substituting these values of x into the second equation, we get y = ±2/√2.
So, we have two more potential points (3/4√2, 2/√2) and (-3/4√2, -2/√2).
Step 4: Evaluate the function at the potential points
Evaluate the function f(x, y) = xy at all the potential points we found.
f(0, -3) = 0 * (-3) = 0
f(0, 3) = 0 * 3 = 0
f(3/4√2, 2/√2) = (3/4√2) * (2/√2) = 3/4
f(-3/4√2, -2/√2) = (-3/4√2) * (-2/√2) = 3/4
Step 5: Compare the function values
Comparing the function values, we see that the maximum value is 3/4 and the minimum value is 0.
Therefore, the maximum value of f(x, y) = xy on the ellipse 8x^2 + y^2 = 9 is 3/4, and the minimum value is 0.