# 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 $$f$$.