EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 11z on the curve of intersection of the plane x − y + z = 1 and the cylinder x2 + y2 = 1. SOLUTION We maximize the function f(x, y, z) = x + 2y + 11z subject to the constraints g(x, y, z) = x − y + z = 1 and h(x, y, z) = x2 + y2 = 1. The Lagrange condition is ∇f = λ∇g + μ∇h, so we solve the equations

Maximum value of f=2.41

Step-by-step explanation:

Lagrange Multipliers

It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.

Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.

for some scalar called the Lagrange multiplier.

For more than one restriction, say g(x,y,z)=0 and h(x,y,z)=0, the Lagrange condition is

The gradient of f is

Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in .

We have

Let's compute the partial derivatives

The Lagrange condition leads to

Operating and simplifying

Replacing the value of in the two first equations, we get

From the first equation

Replacing into the second

Or, equivalently

Squaring

To solve, we use the restriction h

Multiplying by 100

Replacing the above condition

Solving for x

We compute the values of y by solving

For

And for

Finally, we get z using the other restriction

Or:

The first solution yields to

And the second solution gives us

Complete first solution:

Replacing into f, we get

f(x,y,z)=-0.4

Complete second solution:

Replacing into f, we get

f(x,y,z)=2.4

The second solution maximizes f to 2.4

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