Final Answer:
The absolute minima of the function f(x, y) = x² - 2xy + xy^(3/2) on the closed region in the xy-plane bounded below by the parabola y = x², above by the line y = x + 1 is -6/7 and the absolute maxima is 3.
Step-by-step explanation:
In order to find the absolute minimum and maximum of the given function f(x, y) = x² - 2xy + xy^(3/2) on the closed region in the xy-plane bounded below by the parabola y = x², above by the line y = x + 1, we need to find the critical points of the function. The critical points of the function are found by setting the partial derivatives of the function to zero. Therefore, the partial derivatives of the function are as follows:
The partial derivative of f with respect to x (dx) is 2x-2y+xy^(3/2).
The partial derivative of f with respect to y (dy) is -2x+3/2xy^(1/2).
Setting these partial derivatives to zero gives us two equations, which we can solve simultaneously to get the critical points of the given function. Solving these equations simultaneously gives us the critical points (x, y) = (2/3, 4/3) and (-2/3, -4/3).
Now that we have the critical points, we can find the absolute minimum and maximum of the function by plugging these values into the function. Plugging the critical points into the given function gives us f(2/3, 4/3) = -6/7 and f(-2/3, -4/3) = 3. Therefore, the absolute minima of the function f(x, y) = x² - 2xy + xy^(3/2) on the closed region in the xy-plane bounded below by the parabola y = x², above by the line y = x + 1 is -6/7 and the absolute maxima is 3.