15.3 | Constrained Optimization: Lagrange Multipliers


Method of Lagrange Multipliers: To find the maximum and minimum values of \(f(x,y,z)\) subject to the constraint \(g(x,y,z) = k\), assuming that these extreme values exist and \(\nabla g \not = \vec{0} \) on the surface \(g(x,y,z) = k\):

  1. Find all values of \(x,y,z,\) and \(\lambda\) such that
    $$
    \nabla f(x,y,z) = \lambda \nabla g(x,y,z)
    $$
    and
    $$
    g(x,y,z) = k
    $$
  2. Evaluate \(f\) at all the points \((x,y,z)\) that result from step \((1)\). The largest of these values is the maximum value of \(f\); the smallest is the minimum value of \(f\).

Strategy for Optimizing \(f(x,y)\) Subject to the Constraint \(g(x,y) \leq c \)

  • Find all points in the region \(g(x,y) < c\) where \(\nabla f = 0\) or is undefined.
  • Use Lagrange multipiers to find the local extrema of \(f\) on the boundary \(g(x,y) = c\).
  • Evaluate \(f\) at the points found in the previous two steps and compare the values.

Find the maximum and minimum values of
$$
f(x,y) = x^2 + y^2
$$
subject to the constraint \(xy = 1\)


Find all values of \(x,y,z,\) and \(\lambda\) such that
$$
\nabla f (x,y) = \lambda \nabla g(x,y)
$$
where \(g(x,y) = xy = 1\).

That is, solve the system
$$
\left \lbrace
\begin{aligned}
2x & = \lambda y \\
2y & = \lambda x \\
xy & = 1
\end{aligned} \right.
$$

Multiplying the first equation by \(x\) and the second by \(y\) reduces the system to
$$
\left \lbrace
\begin{aligned}
x^2 & = y^2 \\
xy & = 1
\end{aligned} \right.
$$

If \(y = 0\) then \(xy \not = 1\), a contradiction.

So \(y \not = 0\) and therefore we can solve for \(x\) in the second equation. Substituting this back into the first equation, reduces the system to
$$
\frac{1}{y^2} = y^2 \Rightarrow y = \pm 1
$$
Thus the points which satisfy the lagrange conditions are:

  • \((1,1)\)
  • \((-1,-1)\)

Check that this is the case by substituting these points back into the original system. That is, check that \((1,1)\) and \((-1,-1)\) satisfy
$$
\left \lbrace
\begin{aligned}
2x & = \lambda y \\
2y & = \lambda x \\
xy & = 1
\end{aligned} \right.
$$
for some \(\lambda \not = 0\).

Find the maximum and minimum values of
$$
f(x,y,z) = 2x+6y+10z
$$
subject to the constraint \(x^2 + y^2 + z^2 = 35\).


Find all values of \(x,y,z,\) and \(\lambda\) such that
$$
\left \lbrace
\begin{aligned}
\nabla f (x,y) & = \lambda \nabla g(x,y) \\
x^2 + y^2 + z^2 & = 35
\end{aligned} \right.
$$
where \(g(x,y) = x^2 + y^2 + z^2 = 35\).

That is, solve the system
$$
\left \lbrace
\begin{aligned}
2 & = 2\lambda x \\
6 & = 2\lambda y \\
10 & = 2 \lambda z \\
x^2 + y^2 + z^2 & = 35
\end{aligned} \right.
$$

Multiplying the first equation by \(yz\), the second by \(xz\), and the third equation by \(xy\) reduces the system to
$$
\left \lbrace
\begin{aligned}
2yz & = 2\lambda xyz \\
6xz & = 2\lambda xyz \\
10xy & = 2 \lambda xyz \\
x^2 + y^2 + z^2 & = 35
\end{aligned} \right.
$$
which is equivalent to
$$
\left \lbrace
\begin{aligned}
2yz & = 6xz\\
6xz & = 10xy\\
x^2 + y^2 + z^2 & = 35
\end{aligned} \right.
$$
which is equivalent to
$$
\left \lbrace
\begin{aligned}
2z(y- 3x)& =0 \\
6xz & = 10xy\\
x^2 + y^2 + z^2 & = 35
\end{aligned} \right.
$$
That is, either \(z = 0\) or \(y = 3x\).

  • If \(z = 0\), then
    $$
    \left \lbrace
    \begin{aligned}
    0 & = 10xy\\
    x^2 + y^2 & = 35
    \end{aligned} \right.
    $$
    So, either \(x = 0\) or \(y = 0\) but not both since then the second equation won’t be satisfied.

    • If \(x = 0\), then \(y = \pm \sqrt{35}\).
    • If \(y = 0\), then \(x = \pm \sqrt{35}\).

    Therefore \(f\) has possible extreme values at \((\pm \sqrt{35},0,0)\) and \((0,\pm \sqrt{35},0)\).

  • If \(y = 3x\), then
    $$
    \left \lbrace
    \begin{aligned}
    6x(z -5x) & = 0\\
    x^2 + y^2 + z^2 & = 35
    \end{aligned} \right.
    $$
    So, either \(x = 0\) or \(z =5x\).

    • If \(x = 0\), then \(y = 0\) and \(z =\pm \sqrt{35}\).
    • If \(z = 5x\), then
      $$
      x^2 + (3x)^2 + (5x)^2 = 35
      $$
      So, \(35x^2 = 35\). So, \(x = \pm 1\).

    Therefore \(f\) has possible extreme values at \((1,3,5)\) and \((-1,-3,-5)\)

Evaluate \(f\) at these six points, the largest is the maximum value of \(f\) and the smallest is the minimum value of \(f\).

E 15.3 Exercises