Question

Find the absolute maximum and minimum values of f(x,y)=x³+2y²+4 on the set D where D is the closed region bounded by y=0 and y=36−4x²

Part 1: Critical Points The critical points of f are: -
Part 2: Boundary Work The boundary of the region can be expressed by 2 curves

Answer 1

To find the absolute maximum and minimum values of the function f(x, y) = x³ + 2y² + 4 on the region D bounded by y = 0 and y = 36 - 4x², we need to consider both the critical points and the boundary of the region. The critical points are (0, 0), and we can find the maximum and minimum values on the boundary by evaluating the function at the endpoints.

Step-by-step explanation:

To find the absolute maximum and minimum values of the function f(x, y) = x³ + 2y² + 4 on the region D bounded by y = 0 and y = 36 - 4x², we need to consider both the critical points and the boundary of the region.

Part 1: Critical Points

To find the critical points, we need to find the points where the partial derivatives of f with respect to x and y are equal to zero. Taking the partial derivatives, we get:

∂f/∂x = 3x² = 0, ∂f/∂y = 4y = 0

Solving these equations, we find that the critical points are (0, 0).

Part 2: Boundary Work

To find the maximum and minimum values on the boundary, we need to evaluate the function at the endpoints of the boundary. The boundary can be expressed as y = 36 - 4x². Substituting this into the function, we get:

f(x, 36 - 4x²) = x³ + 2(36 - 4x²)² + 4

By analyzing the behavior of the function and the endpoints of the boundary, we can determine the absolute maximum and minimum values.