Question

Create a Lagrange multiplier example with constraints that is unbounded and has a max and a min

Answer 1

a multivariable Greate a Lagranges multiplier example with constraints that is unbounded and has a maximum and minimum Answer: Ket nd, 2 be non-negative real number subject to the condition x+y+z=1. This is consider as our constraints that is unbounded We want to maximize and minimize function f (x, y, z) = xy + y z +Zx-2xyz Subject to the constraints x+y+z=1. Introduce a new varible a, and define a new function Kagrangiam' as follows: Le (n. Y, Z ) = xy + y Z+Z2-2nyz-A (x+y+z-1) To find the critical points of a (xsy, z, d), set the gradient of h equal to zero vector. That is, TA (ASY, Z, A) = 0 (zero vector in 4) ou oy = 0, = 0, 2L Therefore ad a az 2x = 0 = y + z-yz-x=0 2L sy = 0 => x+Z - 2x2 - x=0 and ah az = 0 = x +y-ay-x=0 au = 0 - + y + z-1) = 0 Y+Z-2yz = x+Z-222 = xty-ony = x; and. x+y+z=1 (9-x) (1 - 2Z) = 0 = (2-4) (1-2x) = (x-2) (1-2y) and x+y+z=1

Z=12 This yields the following: g=x or 7=y or a=12 and x+y+z=1 x=z y = Y2 Observe that if none of x,y,z is equal to Y2 , then x=y=Z=12 and f(ny, z) = 6 + 2 + 1 - 2 / 2 27 27 7 and aty x=y So consider the case when t = 12. Then f(x, y, Y2) į [co x+y+z=1 < 7/27 By symmetry, this is case with x=y) as well as with = y=72.

One should remember that the above set of solutions gives all possible local extrema, in the interior of the region under consideration. Since a global extrema which is in the interior of the constraint domain has amongst these but not those which are on the boundary of the constraint domain. So let us look at once of the boundary pieces say given by z=0. Then aty+z= 1 = x+y=1 and flasy, o) = xy Consider L, (xsJ, A) = ny-x, (x+y-1) Putting TL, = 0 y=&=x, = 72 and f (72 72,0) = 44 [7/27. Other contraints such as a=0 or y=0 same result because of symmetry. We have to consider the boundary constraints two at a time, say y=z=0. But then f(0,0,0)=0 27/27 yield the -- x=y=Z=1/3: Iraid, 2) = 7/27 is maximum value and for x = 1, y =O=2: f(x), z = 0 is minimum value. Thus we have an inequality: os xy + y Z+Zx-xyz L 4/27 Whenever x,y,z)0 and aty+z=1.