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

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.

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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

you didn't give me the choices, so i will just give you the graph so you can look at your choices and choose the right one.

jeff jeff

step-by-step explanation: