Question

Find the extremum of f(x, y) subject to the given constraint, and state whether it is a maximum or a minimum.

f(x, y) = 3x²+3y²; 2x+2y = 48
There is a ____ value of ____ located at (x, y) = ____ .

Answer 1

To find the extremum of f(x, y) = 3x²+3y²subject to the given constraint, we can use the method of Lagrange multipliers. The extremum of f(x, y) is a minimum value of 960, located at (x, y) = (4, 20).

Step-by-step explanation:

To find the extremum of f(x, y) = 3x² + 3y² subject to the constraint 2x + 2y = 48, we can use the method of Lagrange multipliers. We define the Lagrangian function as L(x, y, λ) = f(x, y) - λ(g(x, y) - c), where λ is the Lagrange multiplier and c is the constant value of the constraint.

First, we find the gradient of the function and the constraint, which is ∇f(x, y) = (6x, 6y) and ∇g(x, y) = (2, 2) respectively. Setting the partial derivatives equal to zero, we have 6x = 2λ and 6y = 2λ.

By substituting the values of x and y into the constraint equation, we can solve for λ. Solving for x and y gives us x = 4 and y = 20. The extremum of f(x, y) subject to the constraint is a minimum value of 960, located at (x, y) = (4, 20).