Question

Consider the function f(x,y)=x³+6x²y+ 34y³−y on the domain T consisting of the triangle with vertices (−2,−1),(0,1),(2,−1) and the interior of this triangle. We want to find the absolute extrema of the continuous function f on the closed and bounded domain T. (6 points) Find the critical points and singular points of f in the interior of the domain T. Show the steps that detail how you found your critical points. Hint: Solve the system of equations resulting from setting ∂x∂f(x,y) to 0 and ∂y∂f(x,y) equal to 0 .

Answer 1

To find the absolute extrema of the continuous function f(x,y) on the domain T, we must find the critical points by setting the partial derivatives ∂f/∂x and ∂f/∂y to zero and solving the resulting system of equations for x and y.

Step-by-step explanation:

To find the absolute extrema of the function f(x,y)=x³+6x²y+34y³-y within the domain T, we first need to find the critical points in the interior of T. Critical points occur where the partial derivatives of the function with respect to x and y are both zero. In this case, we need to solve the system of equations obtained by setting ∂f/∂x and ∂f/∂y equal to zero.

First, calculate the partial derivative with respect to x, which is ∂f/∂x = 3x² + 12xy. Setting this equal to zero gives us the equation 3x² + 12xy = 0. Similarly, calculate the partial derivative with respect to y, which is ∂f/∂y = 6x² + 102y² - 1. Setting this equal to zero gives us the equation 6x² + 102y² - 1 = 0.

Next, solve the system of equations. From the first equation, we can factor out 3x to get 3x(x + 4y) = 0. This gives us two possible scenarios: either x = 0 or x + 4y = 0. If x = 0, substituting into the second equation gives us 102y² - 1 = 0, which we can solve for y. If x + 4y = 0, we can substitute -4y for x in the second equation and solve for y.

Once y values are obtained, they can be used to find corresponding x values. The pairs (x,y) that lie within the domain T are the critical points in the interior of T.