Homework SourseIf fxy x2 4xy y3 4y find the local and absolute extrema
If f(x,y) = x^2 - 4xy + y^3 + 4y, find the local and absolute extrema (max,min) of f on the triangular region R that has vertices (-1,-1),(7,-1), and (7,7). Can someone show me the steps to solve this thanks!
Solution
f(x,y) = x^2 - 4xy + y^3 + 4y f_x = 2x - 4y f_y = -4x + 3y^2 + 4 for critical points f_x = 0 & f_y = 0 f_x = 0 gives 2x - 4y = 0 x = 2y...........(1) f_y = 0 gives -4x+3y^2 + 4 = 0 3y^2-8y+4 = 0 3y^2-6y-2y+4 = 0 (3y-2)(y-2) = 0 y = 2/3, 2 x = 4/3, 4 hence (4/3, 2/3) & (2,4) are the critical points f_xx = 2 f_yy = 6y at (4/3,2/3) & (2,4) we have minima f(4/3,2/3) = 16/9 - 4*4/3*2/3 + 8/27 + 4*2/3 = 16/9 - 32/9 + 8/27 + 8/3 f(4/3,2/3) = 52/27..........absolute minima f(2,4) = 4 - 4*2*4 + 4^3 + 4*4 = 52.....local minima
