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Next: Curl of a plane Up: Applications of the derivative. Previous: Functions of 3 variables.

   
Lagrange's multipliers.

Proposition 6.25   Let f(x,y,z) be a function and let S be a surface whose equation is given by g(x,y,z)=0. The extrema of f on the surface S are afforded at the points for which there exists a scalar $\lambda$ such that

   \begin{displaymath}
\overrightarrow{\nabla f} = \lambda \overrightarrow{\nabla g}\end{displaymath}

The same proposition is valid for a function of two variables, when the constraint is given by a curve in the plane.

Example 6.26  

Example 6.27   Let f(x,y,z)=x2+2y2+z2 and let S be the cylinder given by x2+4z2=1, i.e. we consider the function g(x,y,z)= x2+4z2-1. Then, we have:

\begin{displaymath}\begin{cases}
\overrightarrow{\nabla f} = 2x \overrightarrow{...
...la g} = 2x \overrightarrow{i} + 8z\overrightarrow{k}\end{cases}\end{displaymath}

We have:

\begin{displaymath}\overrightarrow{\nabla f} = \lambda \overrightarrow{\nabla g}...
...a x \\ 2y =0 \\ 2z= \lambda 8z \\ x^2+y^2+4z^2-1 =0
\end{cases}\end{displaymath}

For which values of $\lambda$ do these equations have a solution?

If x=0, then $z= \pm \frac 12$ and we found two points $P_1(0,0,\frac 12)$ and $P_2(0,0,-\frac 12)$.

If $\lambda =1$, then z=0 and $x=\pm 1$ and this satisfies the first equation too. We get two points P3(1,0,0) and P4(-1,0,0).

Which of these points correpsond to minima and which of them correspond to maxima? We compute the values:

Proposition 6.28   Let f(x,y,z) be a function and let C be the intersection of two surfaces whose equation are given by g1(x,y,z)=0 and g2(x,y,z)=0. The extrema of f on the curve C are afforded at the points for which there exist scalars $\lambda_1$ and $\lambda_2$such that

   \begin{displaymath}
\overrightarrow{\nabla f} = \lambda_1 \overrightarrow{\nabla g_1} +
\lambda_2 \overrightarrow{\nabla g_2}\end{displaymath}

Example 6.29   The plane $\Pi: x-y+x-1=0$ cuts the cylinder x2-y2=1 (displayed on Fig.  7(a)) along a curve $\mathcal{C}$ (one branch is displayed on Fig.  7(b)).
   
Figure 7:
\begin{figure}
\mbox{
\subfigure[]{\epsfig{file=HyperbCylinder.eps,height=6cm} }...
...uad
\subfigure[]{\epsfig{file=HyperbCylinder1.eps,height=6cm} }
}\end{figure}

Find the points on $\mathcal{C}$ at the shortest distance from the origin. This means that we consider the function f(x,y,z)=x2+y2+z2 with the constraints g1(x,y,z)=x-y+x-1 and g2(x,y,z)=x2-y2-1.

We compute the gradients:
\begin{align*}\overrightarrow{\nabla f} &= 2x \overrightarrow{i} + 2y \overright...
...ghtarrow{\nabla g_2} &= 2x \overrightarrow{i} - 2y\overrightarrow{j}\end{align*}
Using Prop.  6.28, we have:

\begin{displaymath}\overrightarrow{\nabla f} = \lambda_1 \overrightarrow{\nabla ...
...lambda_1 - 2y \lambda _2 \\
2z = \lambda_1
\end{cases}Thus,
\end{displaymath}


next up previous contents
Next: Curl of a plane Up: Applications of the derivative. Previous: Functions of 3 variables.
Noah Dana-Picard
2001-05-30