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Lagrange Multipliers

Procedure

Lagrange multipliers are a method used to optimize an objective function when the variables are constrained. Before, we used a constraint to reduce a function of three variables to one of two variables. Now, the function is given in terms of two variables, and x and y are limited.

Goal:

Maximize or minimize an objective function $$f(x,y)$$.

Constraints:

$g(x,y)=0$

Procedure:

Let the objective function $$f$$ and the constraint function $$g$$ be differentiable on a region of $$\mathbb{R}^2$$ with $$\nabla g (x,y) \neq \vec{0}$$ on the curve $$g (x,y) = 0$$.

1. Find $$x$$, $$y$$, and $$\lambda$$ such that $\nabla f (x,y) = \lambda \nabla g (x,y)\ \textrm{ and }\ g(x,y) = 0.$ These are the candidates for the maxima and minima.
2. Select the points $$(x,y)$$ that give the max/min values for $$f(x,y)$$.

Examples

Find the absolute maximum and minimum for $$f(x,y)=2x^2+y^2+2$$ subject to $$g(x,y)=x^2+4y^2-4=0$$.

Solution

\begin{align} \nabla f (x,y) &= 4x \hat{\imath} + 2y \hat{\jmath}\\ \\ \nabla g (x,y) &= 2x \hat{\imath} + 8y \hat{\jmath} \end{align}

Set up the equality:

\begin{align} \nabla f (x,y) &= \lambda \nabla g (x,y)\\ \\ 4x &= \lambda 2x \longrightarrow 2x(2-\lambda)=0 \longrightarrow x=0 \textrm{ or } \lambda = 2\\ \textrm{ and } 2y &= \lambda 8y \longrightarrow 2y(1-4\lambda) = 0 \longrightarrow y=0 \textrm{ or } \lambda = \dfrac{1}{4} \end{align}

The third condition is that $$g(x,y)=0$$:

• If $$x=0$$: $0^2+4y^2-4 = 0 \longrightarrow y=1,-1$
• If $$y=0$$: $x^2+4(0)^2-4 = 0 \longrightarrow x = 2,-2$

Thus, we have four candidates for the maximum and minimum: $$(0,1)$$, $$(0,-1)$$, $$(2,0)$$, and $$(-2,0)$$. We just need to test the function $$f$$ at these four points and see which gives the largest value and which gives the smallest.

$\begin{array}{c | c} (x,y) & f(x,y)\\ \hline (0,1) & 3\\ (0,-1) & 3\\ (2,0) & 10\\ (-2,0) & 10 \end{array}$

Therefore, the maximum of this function is 10, attained at $$(2,0)$$ and $$(-2,0)$$, and the minimum is 3, attained at $$(0,1)$$ and $$(0,-1)$$.

Try it yourself:

(click on a problem to show/hide its answer)

1. Find the absolute maximum and minimum of $$f(x,y)=xy$$ subject to the constraint $$x^2+y^2-8y=9$$.
2. Maximum of 9 at $$(3,3)$$ and $$(-3,-3)$$; minimum of $$-3$$ at $$(-\sqrt{3},\sqrt{3})$$ and $$(\sqrt{3},-\sqrt{3})$$

3. Find the absolute maximum and minimum of $$f(x,y)=y^2-4x^2$$ subject to the constraint $$x^2+2y^2=4$$.
4. Maximum of 2 at $$(0,\sqrt{2})$$ and $$(0,-\sqrt{2})$$; minimum of $$-16$$ at $$(2,0)$$ and $$(-2,0)$$

Find the point on the surface $$4x+y-1=0$$ closest to the point $$(1,2,-3)$$.

Solution

It will be easiest to minimize the squared distance from this point (subject to the constraint that the point lies on the surface):

\begin{align} f(x,y,z) &= (x-1)^2+(y-2)^2+(z+3)^2\\ \\ g(x,y,z) &= 4x+y-1 \end{align} Find the gradients: \begin{align} \nabla f (x,y,z) &= \langle 2x-2, 2y-4, 2z+6 \rangle\\ \\ \nabla g (x,y,z) &= \langle 4,1,0 \rangle \end{align}

Set up the equality, and equate the components:

\begin{align} \nabla f (x,y,z) &= \lambda \nabla g (x,y,z)\\ \\ 2x-2 &= 4\lambda \longrightarrow x=1+2\lambda\\ \textrm{ and } 2y-4 &= \lambda \longrightarrow y = 2+\dfrac{\lambda}{2}\\ \textrm{ and } 2z+6 &= 0 \longrightarrow z=-3 \end{align}

The last condition is that $$g(x,y)=0$$:

\begin{align} 4x+y-1 &= 0\\ 4(1+2\lambda)+\left(2+\dfrac{\lambda}{2}\right)-1 &= 0\\ \dfrac{17\lambda}{2} &= -5\\ \lambda &= -\dfrac{10}{17} \end{align}

Therefore, the point on the given surface that is closest to $$(1,2,-3)$$ is

$\left(-\dfrac{3}{17},\dfrac{29}{17},-3\right)$