3.5 Constrained maxima and minima

Problem: Find the nearest point to $(0,0)$ on the curve $x^2+8xy+7y^2=225$. In other words, find the point on the curve where $x^2+y^2$ is smallest.

This is one example of problems of the following type:

Find maxima and minima of a function $f\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}$ on the curve defined by $g(x,y)=0$.

Two possible methods are as follows.

Answer:

Equation (3) has a simple geometrical interpretation. It says that $(f_x{f}_y)=\lambda (g_x{g}_y)$ for some $\lambda \in ℝ$ (assuming $(f_x{f}_y)\ne (\mathrm{0\hspace{0.33em}\hspace{0.33em}0})$). In other words, $\nabla f$ and $\nabla g$ are parallel. Hence at the points in question, the curves $g(x,y)=0$ and $f(x,y)=c$ (for the appropriate $c$) touch each other. When you think about it, this is what you would expect:

Changing the parameter $c$ moves the curve $f(x,y)=c$ around. In the diagram above, we might start with the value $c^{\prime }$ for $f(x,y)$, and see how far we can increase or decrease $c^{\prime }$. We want to do so in such a way that the curves $f(x,y)=c^{\prime }$ and $g(x,y)=0$ still intersect - the point being that if these two curves intersect, it means that there exist $x,y$ satisfying $g(x,y)=0$ and $f(x,y)=c$ (that is, $c$ is a value of $f(x,y)$ which occurs on the curve $g(x,y)=0$). Suppose we keep on increasing or decreasing $c^{\prime }$ until we reach a value $c$ such that if we carry on moving in the same direction then the curves $f(x,y)=c$ and $g(x,y)=0$ no longer have any points of intersection. This is demonstrated by the curve $f(x,y)=c$ in the diagram above. Then we have pushed (either increased or decreased) $c$ as far as it will go; locally, $c$ is an extreme (either maximum or minimum) value of $f(x,y)$ on the curve $g(x,y)=0$. This is what it means for the curves $f(x,y)=c$ and $g(x,y)=0$ to have a common tangent.

Let’s look a little bit more at the formulation $\nabla f=\lambda \nabla g$. Consider the function:

$$\mathrm{\Lambda }(x,y,\lambda )=f(x,y)-\lambda g(x,y)$$

Think of $\lambda$ as a variable, and let’s try to find maxima and minima of $\mathrm{\Lambda }(x,y,\lambda )$ as $x,y$ and $\lambda$ vary. Then all partial derivates are zero, hence: $f_x-\lambda g_x=f_y-\lambda g_y=0$, and $g(x,y)=0$. The final equation is just the constraint that we want to impose on $x$ and $y$ from the outset; the first two equations together imply $\nabla f=\lambda \nabla g$, which is exactly what we want. This indicates a useful method for finding the (constrained) maxima and minima of $f(x,y)$, by introducing the Lagrange multiplier $\lambda$ and looking for stationary points of the auxiliary function $\mathrm{\Lambda }(x,y,\lambda )$. In a moment we will see some examples that should make things a bit clearer; however, first we will see that a very similar method allows us to find constrained maxima and minima for functions of three or more variables.

Note that this method identifies stationary points, but gives no test to determine whether they are maxima or minima. Just as for a function of one variable, we can find stationary points which are neither maxima nor minima. To see whether a point is a local maximum or a local minimum, you have to use the special features of any particular problem.

In principle, the equation $g(x,y,z)=0$ can be solved to give $z=z(x,y)$. This means that

$$g[x,y,z(x,y)]=0$$

for all $x,y$. Taking partial derivatives with respect to $x$ and $y$ by the chain rule, we obtain

$$g_x+g_z{z}_x=0,g_y+g_z{z}_y=0.$$

(4)

We are looking for stationary points of $f[x,y,z(x,y)]$ (note that this is a function of just $x$ and $y$). At such points, the partial derivatives with respect to $x$ and $y$ will be 0, in other words:

$$f_x+f_z{z}_x=0,f_y+f_z{z}_y=0.$$

(5)

From the $x$-equations in (4) and (5), eliminating the second term as before, we obtain $f_x{g}_z=f_z{g}_x$. From the $y$-equations, we obtain $f_y{g}_z=f_z{g}_y$. Taken together, these two identities say

$$\frac{f_x}{g_x}=\frac{f_y}{g_y}=\frac{f_z}{g_z},$$

and we denote this common value by $\lambda$. We can rewrite this as

$$f_x=\lambda g_x,f_y=\lambda g_y,f_z=\lambda g_z.$$

(6)

or more concisely as $\nabla f=\lambda \nabla g$.

Just as for functions of two variables, we can find the solutions by looking for the stationary points of the auxiliary function $\mathrm{\Lambda }(x,y,z,\lambda )=f(x,y,z)-\lambda g(x,y,z)$.

Let’s now apply this method of Lagrange multipliers to the example given in the beginning of this section.

The auxiliary equation is $\mathrm{\Lambda }(x,y,\lambda )=x^2+y^2-\lambda (x^2+8xy+7y^2-225)$. Now find the partial derivatives:

$${\mathrm{\Lambda }}_x=\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}}{\mathrm{\Lambda }}_y=\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}$$

and ${\mathrm{\Lambda }}_{\lambda }=-(x^2+8xy+7y^2-225)$. So if $(x,y)$ is a stationary point, then

Then, since ${\mathrm{\Lambda }}_{\lambda }=0$, we have $x^2=225$, and hence $x=\pm 15$. However, plugging $(15,0)$ or $(-15,0)$ into ${\mathrm{\Lambda }}_x=0$ gives $\lambda =1$, and into ${\mathrm{\Lambda }}_y=0$ gives $\lambda =0$, a contradiction.

Then $(1-\lambda )x=$

So $y=2x$. Now we can determine possible values of $x$ and $y$ using the final equation ${\mathrm{\Lambda }}_{\lambda }$, that is, $g(x,y)=x^2+8xy+7y^2-225=0$. We obtain

$$

and hence $x=\pm \sqrt{5}$. Thus we obtain two stationary points for $\lambda =1/9$: $(\sqrt{5},2\sqrt{5})$ and $(-\sqrt{5},-2\sqrt{5})$. For both points, $x^2+y^2=25$.

Then $(1-\lambda )x=$

So $x=-2y$. Now $g(x,y)=$

hence $-5y^2=225$, which has no solution.

We have therefore found all of the stationary points of $\mathrm{\Lambda }(x,y,\lambda )$: the two points $\pm (\sqrt{5},2\sqrt{5})$. We don’t yet know whether they are maxima or minima. An ad hoc way to see whether they are maxima or minima is: the curve $x^2+8xy+7y^2-225=0$ passes through the point $(15,0)$, for which $x^2+y^2=225$. Thus $25$ must be the minimum (not maximum) value of $x^2+y^2$. From a strict mathematical standpoint, this argument isn’t really good enough! As mentioned above, not all stationary points are maxima or minima: so perhaps it’s possible that there are some points on the curve $x^2+8xy+y^2=225$ for which $x^2+y^2>25$, and other points for which ? However, for the purposes of this course, such ad hoc arguments will be considered acceptable. (We don’t want you to get bogged down in details of the nature of the stationary points.)

For those who are interested, there is (in this case) an easy trick to be really sure that $25$ is a minimum value: suppose there exist $x,y$ such that $x^2+8xy+7y^2=225$ and . Then . But $9(x^2+y^2)-(x^2+8xy+7y^2)=8x^2-8xy+2y^2=2{(2x-y)}^2\ge 0$, hence this is impossible.

First of all, let’s write the equations down:

$$xyz-4a^3=0,A(x,y,z)=2xy+2xz+yz$$

There are three ways to find the minimum value of $A$.

Since $xyz=4a^3$, we can replace $z$ by $4a^3/xy$, hence $A(x,y,z)=2xy+8a^3/y+4a^3/x$. Now look for the stationary points of $B(x,y)=2xy+8a^3/y+4a^3/x$; we have

hence there is one stationary point: $(a,2a)$. We get a value of $12a^2$ for $B(a,2a)$. To see that this is a minimum value, one could check the double derivatives of $B(x,y)$.

The second method is based on the following useful result:

Proposition: Let $a_1$,$a_2,\mathrm{\dots },a_n$ be positive numbers. Let

Then $G\le A$. Equivalently, $a_1{a}_2\mathrm{\dots }{a}_n\le A^n$.