Using a Lagrange Multiplier with One Constraint.
Find the maximum value of f(x,y)=4xy, where x>0 and y>0, subject to the constraint (x^(2)/3^(2)+(y^(2)/4^(2)=1.
f(3/√2 , 2√2) = 24
To find the maximum value of the function f(x,y) = 4xy with the given constraint, we can use the method of Lagrange multipliers.
Step 1: Write down the function and the constraint
f(x,y) = 4xy
g(x,y) = (x^2/3^2) + (y^2/4^2) - 1
Step 2: Form the Lagrangian function
The Lagrangian function is given by L(x, y, λ) = f(x, y) - λg(x, y), where λ is the Lagrange multiplier.
L(x, y, λ) = 4xy - λ[(x^2/3^2) + (y^2/4^2) - 1]
Step 3: Find the partial derivatives of L(x, y, λ) with respect to x, y, and λ
∂L/∂x = 4y - (2λx/3^2)
∂L/∂y = 4x - (2λy/4^2)
∂L/∂λ = -(x^2/3^2) - (y^2/4^2) + 1
Step 4: Set the partial derivatives equal to zero and solve the system of equations
Setting ∂L/∂x = 0:
4y - (2λx/3^2) = 0 ---- (1)
Setting ∂L/∂y = 0:
4x - (2λy/4^2) = 0 ---- (2)
Setting ∂L/∂λ = 0:
-(x^2/3^2) - (y^2/4^2) + 1 = 0 ---- (3)
Step 5: Solve the system of equations to find the critical points
Solving equations (1) and (2), we get:
2y = (2λx/3^2) ---- (4)
2x = (2λy/4^2) ---- (5)
Substituting equation (4) into equation (5), we have:
2x = (2λx/3^2)(4^2/y)
2x = (16λx/9)(16/y)
2x = (256λx/9y)
y = 128λ/9 ---- (6)
Substituting equation (6) into equation (4), we get:
2(128λ/9) = (2λx/3^2)
256λ/9 = (2λx/9)
x = 128 ---- (7)
Substituting x = 128 into equation (6), we can find the value of y:
y = 128λ/9
y = 128λ/9
Step 6: Substitute the critical points (x, y) into the original function f(x, y)
f(x,y) = 4xy
f(128, 128λ/9) = 4 * 128 * (128λ/9)
f(128, 128λ/9) = 16384λ
Step 7: Find the values of λ that satisfy the constraint (x^2/3^2) + (y^2/4^2) = 1
Substituting x = 128 and y = 128λ/9 into the constraint equation:
(128^2/3^2) + [(128λ/9)^2/4^2] = 1
Simplifying the equation, we have:
16384/9 + (16384λ^2/81)/16 = 1
16384/9 + 16384λ^2/1296 = 1
16384 + 16384λ^2/1296 = 9
16384λ^2/1296 = 9 - 16384
16384λ^2 = (9 - 16384) * 1296
16384λ^2 = - 14087552 * 1296
λ^2 = - 14087552 * 1296 / 16384
λ^2 = - 11094372
Since the square of a real number cannot be negative, this means there are no values of λ that satisfy the constraint.
Step 8: Determine the maximum value of the function f(x, y)
Since we were unable to find any critical points that satisfy the constraint (x^2/3^2) + (y^2/4^2) = 1, we need to consider the boundary of the constraint.
The boundary of the constraint is given by (x^2/3^2) + (y^2/4^2) = 1, which represents an ellipse centered at the origin.
To find the maximum value of f(x, y) on the boundary of the ellipse, we can use the method of optimization on a closed interval. The function f(x, y) = 4xy will be maximized when x and y are at their respective maximum or minimum values on the boundary.
By analyzing the equation (x^2/3^2) + (y^2/4^2) = 1, we can see that the maximum and minimum values for x and y occur at the endpoints of the ellipse.
When x = 3 and y = 0,
f(3, 0) = 4 * 3 * 0 = 0
When x = -3 and y = 0,
f(-3, 0) = 4 * (-3) * 0 = 0
Therefore, the maximum value of f(x, y) = 4xy subject to the constraint (x^2/3^2) + (y^2/4^2) = 1 is 0, which occurs at the points (3, 0) and (-3, 0) on the boundary of the ellipse.