OBJECTIVES

When you finish this section, you should be able to:

1. Use Lagrange multipliers for an optimization problem with one constraint (p. 892)
2. Use Lagrange multipliers for an optimization problem with two constraints (p. 896)

Find the point in the first quadrant on the hyperbola $xy=4$ where the value of $z=12x+3y$ is a minimum. What is the minimum value?

Solution We seek the minimum value of $z=12x+3y$ subject to the condition, or constraint, that $x$ and $y$ satisfy the equation $xy=4$. Since $x>0$ and $y>0,$ we can express this as a problem in one variable by solving $xy=4$ for $y$ and substituting $y=\dfrac{4}{x}$ in the expression $z=12x+3y.$ $$z=12x+3y \underset{\underset{\color{#0066A7}{\hbox{\(y=\dfrac{4}{x}\)}}}{\color{#0066A7}{\uparrow}}}{=}12x+\frac{12}{x} \qquad x>0$$

Then $$\frac{dz}{dx}=12-\frac{12}{x^{2}}=\dfrac{12x^{2}-12}{x^{2}}\qquad x>0$$

The critical numbers of $z$ are $-1$ and $1.$ We exclude $x=-1$, since $x>0$. To examine $x=1,$ we apply the Second Derivative Test. Since $\dfrac{d^{2}z}{dx^{2}}=\dfrac{24}{x^{3}}>0$ for $x=1$, we conclude that $z$ has a minimum when $x=1$ and $y=\dfrac{4}{x} =4$. The minimum value of $z$ is $24$ at the point $(1,4)$ on the hyperbola.

NEED TO REVIEW?

The Second Derivative Test is discussed in Section 4.4, pp. 291-292.

THEOREM Lagrange's Theorem

Let $f$ and $g$ be functions of two variables with continuous partial derivatives at every point of some open set containing the smooth curve $g(x,y)=0$. Suppose that $f$, when restricted to points on the curve $g(x,y)=0$, has a local extremum at the point $(x_{0},y_{0})$ and that ${\nabla }g(x_{0},y_{0})\neq \mathbf{0}$. Then there is a number $\lambda$ for which $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED]{ {\nabla }f(x_{0},y_{0})=\lambda {\nabla }g(x_{0},y_{0}) }}$$

Proof

Express the smooth curve $C$ defined by $g(x,y)=0$ as a parametric equation in vector form $\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}$. Define the function $s(t)=f(x(t),y(t))$ and choose $t_{0}$ so that $x_{0}=x(t_{0})$, $y_{0}=y(t_{0})$. Now since $f$ has a local extremum at the point $(x_{0},y_{0})$, the extreme value is given by $$\begin{equation*} s(t_{0})=f(x(t_{0}),y(t_{0}))=f( x_{0},y_{0}) \end{equation*}$$

Then $s' (t_{0})=0$. By Chain Rule I, $$\begin{eqnarray*} s' (t_{0}) &=&f_{x}(x_{0},y_{0})x'(t_{0})+f_{y}(x_{0},y_{0})y' (t_{0}) \qquad {\color{#0066A7}{\hbox{\(\tfrac{ds}{dt}=\tfrac{\partial s}{\partial x} \tfrac{dx}{dt}+\tfrac{\partial s}{\partial y}\tfrac{dy}{dt}\)}}}\\ &=&{\nabla }f(x_{0},y_{0})\,{\cdot}\, \mathbf{r}^{\prime }(t_{0}) \qquad\qquad\qquad \enspace \enspace \enspace {\color{#0066A7}{\hbox{Definition of the gradient}}}\\ &=&0 \end{eqnarray*}$$

If ${\nabla }f(x_{0},y_{0})\neq \mathbf{0}$, it is orthogonal to $\mathbf{r}' (t_{0})$, which is a tangent vector to $C$ at $(x_{0},y_{0})$. Since the gradient is normal to the level curve at $( x_{0},y_{0}) ,$ the vector ${\nabla }g(x_{0},y_{0})$ is also orthogonal to $\mathbf{r}' (t_{0})$. If ${\nabla }f(x_{0},y_{0})=\mathbf{0}$, then let $\lambda =0$. Thus, in every case, there is a number $\lambda$ for which ${\nabla }f(x_{0},y_{0})=\lambda {\nabla } g(x_{0},y_{0})$.

ORIGINS

Joseph-Louis Lagrange (1736–1813), a.k.a. Giuseppe Lodovico Lagrangia, was a self-taught mathematician and astronomer. He was born and began his career teaching in Turin, Italy, but later moved first to Berlin and then to Paris, where he held prestigious positions at the Berlin Academy and Acad\smale mie des Sciences. After the French Revolution (1789), Lagrange was named the first professor of analysis at the École Polytechnique, in Paris. Lagrange is one of 72 names of mathematicians, engineers, and scientists inscribed on the Eiffel Tower.

Find the point in the first quadrant on the hyperbola $xy=4,$ where the value of $z=12x+3y$ is a minimum. What is the minimum value?

Solution We can reword the problem to read as follows: Find the minimum value of $z=f(x,y)=12x+3y$ subject to the condition $xy=4,$ where $x>0$ and $y>0$.

The test points satisfy the equations $${\nabla }f(x,y)=\lambda {\nabla }g(x,y)\qquad g(x,y)=xy-4=0\qquad x>0\quad y>0$$

where $\lambda$ is a number.

Solve the system of equations $$\left\{\begin{array}{l} 12 &=&\lambda y \qquad \enspace{\color{#0066A7}{\hbox{\(f_{x} (x,y)=\lambda g_{x} (x,y)\)}}}\\ 3 &=&\lambda x \qquad \enspace {\color{#0066A7}{\hbox{\(f_{y} (x,y)=\lambda g_{y} (x,y)\)}}}\\ xy-4 &=&0 \qquad \enspace \enspace {\color{#0066A7}{\hspace{.3pc}\hbox{\(g(x,y)=0\)}}} \end{array}\right.$$

Eliminating $\lambda$ from the first two equations results in $y=4x$, so the third equation becomes $$\begin{eqnarray*} &\qquad\qquad \enspace 4x^{2}-4=0 \\[2pt] &x=1 \qquad \enspace \enspace \hbox{or} \enspace \enspace \enspace x=-1 \end{eqnarray*}$$

Since $x>0$, we ignore $x=-1.$ When $x=1$, then $y=4$, so the only test point is $(1,4)$. The corresponding minimum value of $z=12x+3y$ is $z=24$.

NOW WORK

Problem 5.

Steps for Using Lagrange Multipliers

Step 1 Express the problem in the following form: Find the maximum (or minimum) value of $z=f( x,y)$ subject to the constraint $g( x,y) =0.$

Step 2 Check that the functions $f$ and $g$ satisfy Lagrange's Theorem.

Step 3 Solve the equations ${\nabla }f( x,y) =\lambda {\nabla }g( x,y)$ and $g( x,y) =0$ for the test points by solving the system of equations $$\left\{\begin{array}{rcl} f_{x}( x,y) &=&\lambda g_{x}( x,y)\\ f_{y}( x,y) &=&\lambda g_{y}( x,y)\\ g( x,y) &=&0\end{array}\right.$$

Step 4 Evaluate $z=f( x,y)$ at each test point found in Step 3. Choose the maximum (or minimum) value of $z.$

Find the maximum and minimum values of $z=f(x,y)=3x-y+1$ subject to the constraint $3x^{2}+y^{2}=9$.

Solution We use the steps for Lagrange multipliers:

Step 1 The problem is expressed in the required form.

Step 2 The functions $f$ and $g( x,y) =3x^{2}+y^{2}-9=0$ each have continuous partial derivatives.

Step 3 Solve the system of equations $$\left\{\begin{array}{rcl} 3&=&6\lambda x & \qquad {\color{#0066A7}{\hbox{\(f_{x}(x,y) =\lambda g_{x}(x,y)\)}}}\\ -1&=&2\lambda y & \qquad {\color{#0066A7}{\hbox{\(f_{y}(x,y) =\lambda g_{y}(x,y)\)}}}\\ 3x^{2}+y^{2}-9&=&0 & \qquad {\color{#0066A7}{\hspace{.3pc}\hbox{\(g(x,y) =0\)}}} \end{array}\right.$$

From the first two equations, we find that $x=\dfrac{1}{2\lambda }$ and $y=-\dfrac{1}{2\lambda }$ from which $x=-y$. Substituting into the third equation, we have $$\begin{eqnarray*} 4x^2&=&9\\[4pt] x&=&\pm \dfrac{3}{2} \end{eqnarray*}$$

Then $x=\pm \dfrac{3}{2},$ $y=\mp \dfrac{3}{2}$, and the test points are $\left( \dfrac{3}{2},-\dfrac{3}{2}\right)$ and $\left( -\dfrac{3}{2},\dfrac{3}{2}\right)$.

Step 4 The values of $z$ corresponding to the test points are $z=\dfrac{9}{2}+\dfrac{3}{2}+1=7$ and $z=-\dfrac{9}{2}-\dfrac{3}{2}+1=-5$. The maximum value of $z$ is $7$ and the minimum value is $-5$.

NOW WORK

Problem 9.

Find the absolute maximum and absolute minimum of $f(x,y)=x^{2}+y^{2}+4x-4y+3$ subject to $x^{2}+y^{2}\leq 2$.

Solution From the Extreme Value Theorem, $f$ has both an absolute maximum and an absolute minimum on the closed, bounded set $x^{2}+y^{2}\leq 2.$ We begin by finding any critical points. $$\begin{equation*} \begin{array}{rcl@{\qquad}rrcl} f_{x}=2x+4&=&0 & f_{y}=2y-4&=&0\\ x&=&-2 & y&=&2 \end{array}\end{equation*}$$

The point $( -2,2)$ lies outside the disk $x^{2}+y^{2}\leq 2,$ so $f$ has no relevant critical points. The extrema must occur on the boundary of the domain, that is, on the curve $x^{2}+y^{2}=2$. To find the extrema, we use Lagrange multipliers.

Step 1 Using $g(x,y)=x^{2}+y^{2}-2=0$ as the constraint, the problem is expressed in the required form.

Step 2 The functions $f$ and $g( x,y) =x^{2}+y^{2}-2=0$ each have continuous partial derivatives.

Step 3 We solve the system of equations ${\nabla }f( x,y) =\lambda {\nabla }g( x,y)$ and $g( x,y) =0.$ $$\left\{ \begin{array}{rcl@{\qquad}l} 2x+4&=&2\lambda x \qquad \enspace{\color{#0066A7}{\hbox{\(f_{x}(x,y) =\lambda g_{x}(x,y)\)}}}\\ 2y-4&=&2\lambda y \qquad \enspace{\color{#0066A7}{\hbox{\(f_{y}(x,y) =\lambda g_{y}(x,y)\)}}}\\ x^{2}+y^{2}&=&2 \qquad \enspace \enspace \enspace \enspace {\color{#0066A7}{\hbox{\(g(x,y) =0\)}}} \end{array}\right.$$

We eliminate $\lambda$ from the first two equations. $$\begin{eqnarray*} 2x+4&=&2x\lambda \qquad \enspace 2y-4&=&2y\lambda\\ \lambda &=& \dfrac{2x+4}{2x} & \lambda &=&\dfrac{2y-4}{2y}\\ \lambda &=& 1+\dfrac{2}{x} & \lambda &=& 1-\dfrac{2}{y} \end{eqnarray*}$$

Then $1+\dfrac{2}{x}=1-\dfrac{2}{y}$, so $y=-x.$

If we substitute $y=-x$ into the third equation $x^{2}+y^{2}=2,$ we find $2x^{2}=2.$ Then $x=\pm 1$ and $y=\mp 1.$ There are two test points: $(1,-1)$ and $(-1,1)$.

Figure 22

Step 4 We evaluate $z=f( x,y)$ at each test point. $$\begin{eqnarray*} f(1,-1)&=&13 \qquad {\color{#0066A7}{\hbox{Maximum value}}}\\ f(-1,1)&=&-3 \qquad {\color{#0066A7}{\hbox{Minimum value}}} \end{eqnarray*}$$

NOW WORK

Problem 17.

A container in the shape of a rectangular box is open on top and has a fixed volume of 12 m$^3$. The material used to make the bottom of the container costs $3 per square meter, while the material used for the sides costs $1 per square meter. What dimensions will minimize the cost of material?

Solution Let $x$ be the width of the container, $y$ its depth, and $z$ its height. Then its volume is $xyz=12.$ The cost function $C=C(x,y,z)$, which is to be minimized, is $$C(x,y,z) =3xy+2xz+2yz$$

We use Lagrange multipliers.

Step 1 The objective is to minimize $C$ subject to the constraint $g(x,y,z)=xyz-12=0$.

Step 2 The functions $C$ and $g$ each have continuous partial derivatives.

Step 3 We solve the system of equations ${\nabla }C(x,y,z) =\lambda {\nabla }g(x,y,z)$ and $g(x,y,z) =0$. $$\left\{\begin{array}{lllll} 3y+2z &=&\lambda yz \qquad {\color{#0066A7}{\hbox{\(C_{x}(x,y,z) =\lambda g_{x}( x,y,z)\)}}} \enspace \enspace {\color{#0066A7}{\hbox{(\(1\))}}}\\ 3x+2z &=&\lambda xz \qquad {\color{#0066A7}{\hbox{\(C_{y}(x,y,z) =\lambda g_{y}( x,y,z)\)}}} \enspace \enspace {\color{#0066A7}{\hbox{(\(2\))}}}\\ 2x+2y &=&\lambda xy \qquad {\color{#0066A7}{\hbox{\(C_{z}(x,y,z) =\lambda g_{z}( x,y,z)\)}}} \enspace \enspace {\color{#0066A7}{\hbox{(\(3\))}}}\\ xyz-12 &=&0 \qquad \enspace \enspace \enspace {\color{#0066A7}{\hbox{\(g( x,y,z) =0\)}}}\quad \enspace \qquad \enspace \enspace \enspace{\color{#0066A7}{\hbox{(\(4\))}}} \end{array}\right.$$

Now we use the facts that $x>0$, $y>0$, and $z>0$ to solve the first three equations for $\lambda$, obtaining $$\lambda =\dfrac{3}{z}+\frac{2}{y}\enspace {\color{#0066A7}{\hbox{(\(5\))}}} \qquad \lambda =\frac{3}{z}+\frac{2}{x}\enspace {\color{#0066A7}{\hbox{(\(6\))}}} \qquad \lambda =\dfrac{2}{y}+\frac{2}{x}\enspace {\color{#0066A7}{\hbox{(\(7\))}}}$$

From these, we find that $$\begin{eqnarray*} y &=&x \qquad \enspace\enspace\enspace {\color{#0066A7}{\hbox{From equations (5) and (6)}}}\\ z &=&\dfrac{3}{2}x \qquad \enspace \enspace {\color{#0066A7}{\hbox{From equations (5) and (7)}}} \end{eqnarray*}$$

If we substitute these two equations into $xyz-12=0$, we find \[ \begin{eqnarray*} xyz-12& =&0 \\[4pt] x( x) \left( \dfrac{3}{2}x\right) -12& =&0 \\[4pt] \frac{3}{2}x^{3}& =&12 \\[4pt] x& =&2 \end{eqnarray*}

Then $y=2,$ $z=3,$ and the test point is $(2,2,3)$.

Step 4 The dimensions of the container that minimize the cost are 2 m by 2 m by 3 m. The minimum cost is $C(2,2,3)=3 ( 2) ( 2) +2 ( 2) (3) +2 ( 2) ( 3) =$36.$

NOW WORK

Problem 33.

Figure 23

Find the points of intersection of the ellipsoid $x^{2}+y^{2}+9z^{2}=25$ and the plane $x+3y-2z=0$ that are farthest from the origin. Also find the points that are closest to the origin.

Solution See Figure 23.

Let $C$ be the curve formed by intersecting the ellipsoid and the plane, and let $P=(x,y,z)$ be a point on $C$. We use the steps for Lagrange multipliers.

Step 1 The function to be optimized is the distance $d$ from $P$ to the origin: $$d=f(x,y,z)=\sqrt{x^{2}+y^{2}+z^{2}}$$

subject to the constraints $$\begin{eqnarray*} g(x,y,z) &=&x^{2}+y^{2}+9z^{2}-25=0 \qquad \enspace \enspace {\color{#0066A7}{\hbox{The point must be on the ellipsoid.}}}\\ h(x,y,z) &=&x+3y-2z=0 \qquad \qquad \qquad {\color{#0066A7}{\hbox{The point must be on the plane.}}} \end{eqnarray*}$$

Step 2 The functions $f,$ $g,$ and $h$ are each functions of three variables, and each has continuous partial derivatives at every point of some open set containing the smooth curve defined by $g(x,y,z)=0$ and $h(x,y,z)=0.$

Step 3 We introduce two Lagrange multipliers and solve the system of equations ${\nabla }f(x,y,z)=\lambda _{1}{\nabla }g(x,y,z)+\lambda _{2}{\nabla }h(x,y,z)$, $g(x,y,z)=0,$ and $h(x,y,z)=0.$

Eliminating $\lambda _{1}$ and $\lambda _{2}$ from the first three equations, we have two possibilities: $z=0$ or $y=3x$. Using each of these equations with the constraints

$x^{2}+y^{2}+9z^{2}- 25=0$ and $x+3y-2z=0$ results in the following: $$\begin{eqnarray*} z&=&0{:} & \left( -\dfrac{15}{\sqrt{10}},\dfrac{5}{\sqrt{10}},0\right) \qquad \qquad \enspace \enspace \enspace {\color{#0066A7}{\hbox{5 units from the origin}}}\\ &&& \left( \dfrac{15}{\sqrt{10}},-\dfrac{5}{\sqrt{10}},0\right) \qquad \qquad \enspace \enspace \enspace {\color{#0066A7}{\hbox{5 units from the origin}}}\\ y&=&3x{:} & \left( \dfrac{5}{\sqrt{235}},\dfrac{15}{\sqrt{235}},\dfrac{25}{\sqrt{235}}\right) \qquad \enspace {\color{#0066A7}{\hbox{\(\sqrt{\dfrac{875}{235}}\approx 1.930\) units from the origin}}}\\ &&& \left( -\dfrac{5}{\sqrt{235}},-\dfrac{15}{\sqrt{235}},-\dfrac{25}{\sqrt{235}}\right) \enspace \enspace {\color{#0066A7}{\hbox{\(\sqrt{\dfrac{875}{235}}\approx\) 1.930 units from the origin}}} \end{eqnarray*}$$

The points $\left( -\dfrac{15}{\sqrt{10}},\dfrac{5}{\sqrt{10}},0\right)$ and $\left( \dfrac{15}{\sqrt{10}},-\dfrac{5}{\sqrt{10}},0\right)$ are farthest from the origin, and the points $\left( \dfrac{5}{\sqrt{235}},\dfrac{15}{\sqrt{235}},\dfrac{25}{\sqrt{235}}\right)$ and $\left( -\dfrac{5}{\sqrt{235}},-\dfrac{15}{\sqrt{235}},-\dfrac{25}{\sqrt{235}}\right)$ are closest to the origin. Since the intersection of the ellipsoid and the plane is an ellipse, the points we have found are the endpoints of the major and minor axes of this ellipse.

NOW WORK

Problem 19.