16. Max-Min Problems

b. Constrained Max-Min Problems

2. Lagrange Multiplier Method

Since the maximum and minimum of a function \(f\) on a constraint \(g=C\) occur at points where the level curve of \(f\) is tangent to the constraint, which is itself a level curve of \(g\). Since the level curves are tangent, their normals are parallel. Further, the normals are the gradients of the functions \(f\) and \(g\). However, although the gradients \(\vec\nabla f\) and \(\vec\nabla g\) are parallel, their lengths can be different, i.e.: \[ \vec\nabla f=\lambda\vec\nabla g \] for some number \(\lambda\) (the Greek letter lambda) which is called the Lagrange multiplier. These equations are called the Lagrange equations. Since \(f\) and \(g\) are functions of \(2\) variables, there are \(2\) components in the gradients (the \(2\) partial derivatives) and so the Lagrange equations provide \(2\) equations for the \(3\) variables \(x\), \(y\) and \(\lambda\). We need one more equation. It is the constraint equation \(g=C\). So we solve the Lagrange equations together with the constraint for \(x\), \(y\) and \(\lambda\). Similar results hold for functions of more variables. (See below.)

Although in principal, we are solving for \(\lambda\). In fact, we don't really care about its value. It is merely a mathematical construct that helps us solve for the extrema.

Let's start with an example with only \(2\) variables which could be solved by the methods used in single-variable calculus, although with a lot more work.

Find the maximum and minimum values of the function \(f=x^2+y^2\) on the ellipse \(\dfrac{(x+y)^2}{16}+\dfrac{(x-y)^2}{9}=1\).

Since we are solving this equation using the Lagrange multiplier method, the first thing we need are the gradients of the two functions: \[\begin{aligned} \vec{\nabla}f&=\langle 2x,2y\rangle \\ \vec{\nabla}g&=\left\langle \dfrac{2(x+y) }{16}+\dfrac{2(x-y)}{9}, \dfrac{2(x+y) }{16}+\dfrac{-2(x-y) }{9}\right\rangle \end{aligned}\] We now insert these gradients into the Lagrange equations, \[ \vec{\nabla}f=\lambda\vec{\nabla}g \] The \(x\) and \(y\) components are: \[\begin{aligned} 2x&=\lambda \left(\dfrac{2(x+y)}{16}+\dfrac{2(x-y)}{9}\right) \qquad \text{(1)} \\ 2y&=\lambda \left(\dfrac{2(x+y)}{16}-\,\dfrac{2(x-y)}{9}\right) \qquad \text{(2)} \end{aligned}\] This is \(2\) equations, but there are \(3\) unknowns, \(x\), \(y\) and \(\lambda\). So we need a third equation. It is the constraint equation: \[ \dfrac{(x+y)^2}{16}+\dfrac{(x-y)^2}{9}=1 \qquad \text{(3)} \] To solve these, we divide equations (1) and (2) by \(2\) and then add and subtract them: \[\begin{aligned} x+y&=\lambda\dfrac{2(x+y)}{16} \qquad \text{(5)} \\ x-y&=\lambda\dfrac{2(x-y)}{9} \qquad \text{(6)} \end{aligned}\] We now rearrange (5) and (6) to factor out \(x+y\) and \(x-y\): \[\begin{aligned} (x+y)\left(1-\,\dfrac{\lambda}{8}\right)=0 \qquad \text{(7)} \\ (x-y)\left(1-\,\dfrac{2\lambda}{9}\right)=0 \qquad \text{(8)} \end{aligned}\] Equation (7) gives the cases \(x+y=0\) or \(\lambda=8\). Equation (8) gives the cases \(x-y=0\) or \(\lambda=\dfrac{9}{2}\). Together, these give only two cases:

• Case 1: \(\lambda=\dfrac{9}{2}\) and \(x+y=0\):
  With \(y=-x\), the constraint equation (3) gives: \[ (2x)^2=9 \qquad \text{or} \qquad x=\pm\dfrac{3}{2} \] So the critical points are \(\left(\dfrac{3}{2},-\,\dfrac{3}{2}\right)\) and \(\left(-\,\dfrac{3}{2},\dfrac{3}{2}\right)\). At both of these points the function value is: \[ f=x^2+y^2=\dfrac{9}{4}+\dfrac{9}{4}=\dfrac{9}{2} \]
• Case 2: \(\lambda=8\) and \(x-y=0\):
  With \(y=x\), the constraint equation (3) gives: \[ (2x)^2=16 \qquad \text{or} \qquad x=\pm2 \] So the critical points are \((2,2)\) and \((-2,-2)\). At both of these points the function value is: \[ f=x^2+y^2=4+4=8 \]

So the maximum is \(f=8\) at \((2,2)\) and \((-2,-2)\) and the minimum is \(f=\dfrac{9}{2}\) at \(\left(\dfrac{3}{2},-\,\dfrac{3}{2}\right)\) and \(\left(-\,\dfrac{3}{2},\dfrac{3}{2}\right)\). Check these are reasonable in the plot. Notice that we don't care about the values of \(\lambda\).

Notice that in a problem like this we do not need to use the Second Derivative Test because we know we have found all the critical points. The one with the largest function value must be the maximum and the one with the smallest function value must be the minimum.

More Variables

The same works for any number of variables. The maximum and minimum of a function \(f\) on a constraint \(g=C\) occur at points where the level set (surface or higher dimensional surface) of \(f\) is tangent to the constraint, which is a level set of \(g\). Since the level sets are tangent, their normals are parallel. Further, the normals are the gradients of the functions \(f\) and \(g\). Once again, although the gradients \(\vec\nabla f\) and \(\vec\nabla g\) are parallel, their lengths can be different, i.e.: \[ \vec\nabla f=\lambda\vec\nabla g \] for some Lagrange multiplier \(\lambda\). If \(f\) and \(g\) are functions of \(n\) variables, namely \(x_1, x_2, \cdots, x_n\), then there are \(n\) components in the gradients (the \(n\) partial derivatives) and so the Lagrange equations provide \(n\) equations for the \(n+1\) unknowns \(x_1, x_2, \cdots, x_n\) and the Lagrange multiplier \(\lambda\). We need one more equation. It is the constraint equation \(g=C\). We solve the Lagrange equations together with the constraint. Once again, we don't really care about the value of \(\lambda\). It is merely a mathematical construct that helps us solve for the extrema.

We now redo an exercise from the previous page which was difficult because the derivatives involved square roots. The Lagrange multiplier method avoids the square roots.

From this exercise, you should notice that the Lagrange multiplier method is much easier than eliminating a variable, about half as much work.