Use lagrange multipliers to find the maximum and the minimum values of the function f(x,y,z)=yz+xy,subject the constraints xy=1
and y2+z2 =1
1 answer:
You can use Lagrange multipliers, but you don't need to in order to find the function's extremes. The first constraint tells you
... f(x, y, z) = yz + 1
The second constraint tells you
... f(x, y, z) = y·(±√(1-y²)) +1
A graph of this function shows it to have
... a minimum of 0.5
... a maximum of 1.5.
_____
The Lagrangian can be written as
... L = xy +yz +α(xy -1) +β(y² +z² -1)
Differentiating and setting partial derivatives to zero gives the 5 equations
... ∂L/∂x = 0 = y +αy
... ∂L/∂y = 0 = x + z +αx +2βy
... ∂L/∂z = 0 = y +2βz
... ∂L/∂α = 0 = xy -1
... ∂L/∂β = 0 = y² +z² -1
The first equation tells you α = -1.
Substituting that into the second equation gives one that is symmetrical with the third equation:
... z + 2βy = 0
... y + 2βz = 0
Adding these two equations gives
... (y+z)(1 +2β) = 0
which has solutions
... z = -y ... and ... β = -1/2
Using the latter in either the second or third equation gives
... z = y
Plugging these into the second constraint (the 5th equation), you get
... y² = z² = 1/2
Since z = ±y, yz = ±y² = ±1/2 and the extremes of the objective function become
... f(x, y, z) = ±1/2 + 1 = {1/2, 3/2}
That is ...
... f(-√2, -1/√2, -1/√2) = 3/2 . . . maximum
... f(-√2, -1/√2, 1/√2 = 1/2 . . . . . minimum
... f(√2, 1/√2, 1/√2) = 3/2 . . . . . maximum
... f(√2, 1/√2, -1/√2) = 1/2 . . . . . minimum
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