Finding maximum on curve

Question

Find the maximum of f(x,y)=xy on the constraint curve (x+1)2+y2=1.

Answer 1

Constrained optimization means we'll need to use the Lagrange multiplier method:

 

∇f=λ∇g, 

 

where f is the function to be optimized (xy), and g is a level set of the constraint function (basically, set the constraint =0 and call that g. Here it's g=(x+1)²+y²-1).

 

∇f = [y, x]

 

∇g = [2(x+1), 2y]

 

Plugging these vectors into the Lagrange multiplier equation will set up a system of equations (we set each x-component equal to each other, then each y-component, and so on if you're dealing with longer vectors in more variables]

 

y=2λ(x+1)

x=2λy

 

These are two equations in three variables (x,y,λ). We therefore need a third equation to have a solution to the system. The last equation is the always constraint function itself. The answer to the optimization problem is all coordinate pairs (x,y) that simultaneously satisfy

 

y=2λ(x+1)

x=2λy

(x+1)²+y²=1

 

 

You won't ever see the λ appear in the solution to these types of problems, but it is usually necessary to calculate its value to actually solve the system of equations.

 

From eqn 2: 

λ=x/2y

 

Substitute this into eqn 1:

y=2(x/2y)(x+1)

y²=x(x+1)

 

Substitute this into eqn 3:

(x+1)²+x(x+1)=1

x²+2x+1+x²+x=1

2x²+3x=0

x={0, -3/2}

 

When x=0, y²=0(0+1)=0. One solution is (0,0).

When x=-3/2, y²=(-3/2)(-1/2)=3/4. The other solutions are (-3/2, √[3/4]) and (-3/2, -√[3/4]).

 

These are analogous to critical points, i.e. each could be a maximum or a minimum. Plugging the points in to the original f(x,y) to calculate their values will let you determine which types of extrema you have.