Use the method of Lagrange multipliers to it\": y) = V 332 +92 subject to the constraint 3+1; = 3. A rectangular box with edges parallel to the axes is inscribed in the ellipsoid 1. volume = 27 cu. units 912 + 3y2 +922 = 1 2. volume = |00 cu. units 81 similar to the one shown in 3. volume = 27 cu. units O 8 4. volume cu. units 27 O 5. volume = cu. units 81 Use Lagrange multipliers to determine the maximum volume of this box. Note: all 8 vertices of the box will lie on the ellipsoid when the volume is marimized.Use Lagrange multipliers to find the max- 1. fmax = 256, fmin = -256 imum and minimum values of the function f(x, y) = 4x y, subject to the constraint 2. fmax = 1024, fmin = 0 4x2 + y2 = 192. O 3. fmax = 0, fmin =-512 O 4. fmax = 512, fmin = -512 5. fmax = 1024, fmin = -1024Finding the minimum value of f(a:, y) = 2x+y+ 1 subject to the constraint 9(3, y) = 412+3y23 = 0 is equivalent to nding the height of the lowest point on the curve of intersection of the graphs of f and 9 shown in Use Lagrange multipliers to determine this minimum value