a. Find the critical point of the given function and then determine whether it is a local maximum, local minimum, or saddle point.f(x, y) = x2 − y2 + 3xyb. Find the critical point of the given function and then determine whether it is a local maximum, local minimum, or saddle point.f(x, y) = x2 + y2 − xy
Accepted Solution
A:
Answer:Step-by-step explanation:a) f(x,y) = x² - y² + 3xyFirst derivative x (variable) First derivative y (variable)f´(x) = 2x + 3y f´(y) = - 2y + 3xSecond derivative x (variable) Second derivative y (variable)f´´(xx) = 2 f´´ (yy) = -2Cross derivative Cross derivativef´´(x,y ) = 3 f´´(y,x) = 3f´(x) = 0 and f´(y) = 02x + 3y = 0 - 2y + 3x = 0We got a two equation system with two uknown variables2x + 3y = 0 - 2y + 3x = 0Solvingx = 3/2 * y -2y + 3* (3/2) *y =0-2y + 9/2 y = 0 y = 0 and x = 0 Critical point P ( 0 ,0 )we must evaluate f´´(x,x) = 2f´´(y,y) = -2f´´ (x,y) = f´´(y,x) = 3We compute discriminitingD = f´´(x,x) * f´´(y,y) - [f´´(x,y)]² D = (2)*(-2) - (3)² D = -4 -9D = -13Then we got one second derivative positive the other negative and D < 0we have a saddle pointb) f(x,y) = x² + y² - xyfolowing the same procedure f¨(x) = 2x -y f´(y) = 2y - xf´´ (x,x) = 2 f´´(y,y) = 2Croos derivativef´´(x,y) = -1 f´´(y,x) = -1Equation system: 2x -y = 0 y = 2x 2y - x = 0 2(2x) - x = 0 4x - x = 0 x = 0 and y = 0Critical point P ( 0 , 0 )f´´(x,x) = 2f´´(y,y) = 2 f´´ (x,y) = f´´(y,x) = -1D = (2)*2 - (-1) = 4 - 1 = 3D = 3The fuction has a minimum the point P ( 0 , 0) is a minimum