16.3 Lagrange multipliers

Interpret $f(x,y)$ as a distance squared, and sketch $xy=1$ in the $xy$-plane. You might also want to review section 16.3.3 in the text.

(a) $f$ does not have a maximum. It does have a minimum.

(b) The minima are at $\pm (1,1)$, where $f$ takes the value $2$.

(a) $f(x, y) = x^2 +y^2$ is the square of the distance from the point $(x,y)$ to the origin. There are points on the curve $xy=1$ that have either $x$ or $y$ arbitrarily large and so whose distance from the origin is arbitrarily large. So $f$ has no maximum on the curve.

On the other hand $f$ will have a minimum, achieved at the points of $xy=1$ that are closest to the origin.

(b) On the curve $xy=1$ we have $y=\frac{1}{x}$ and hence $f=x^2+\frac{1}{x^2}$. As

and as no point of the curve has $x=0$, the minimum is achieved when $x=\pm 1$. So the minima are at $\pm (1,1)$, where $f$ takes the value $2$.

Remark: this is less a question specifically about Lagrange multipliers and more a question about the existence of extrema on unbounded curves, as in section 16.3.3 in the text.

The easiest way out is to find a function $z=k(x)$ with local but not absolute extrema, then affix that to the plane $y=0$.

One possible answer: $g(x,y)=y$, $f(x,y)=x^3-x$.

The easiest (cheapest?) way out is to think of a function $z=k(x)$ with local but not absolute extrema, then consider the constraint $y=0$. This puts our function in the $xz$-plane, effectively making it look just like the function of one-variable $y=f(x)$.

For example, we can set $f(x,y)=x^3-x$, with constraint function $g(x,y)=y=0$.

Using techniques from last semester, the function $z=x^3-x$ has local max at $x=-\frac1{\sqrt3}$ and local min at $x=\frac1{\sqrt3}$; but it has no absolute extrema because $\lim\limits_{x \to \infty}(x^3-x)=\infty$ and $\lim\limits_{x \to -\infty}(x^3-x)=-\infty$.

Similarly, $f(x,y)$ has a local constrained max resp. min at $\left(-\frac1{\sqrt3},0\right)$ resp. $\left(-\frac1{\sqrt3},0\right)$; but has no absolute extrema.

Not much calculation is necessary.

There are none

There are none.

For any integer $n$, $\sin\left(\frac{\pi}{2}+2\pi n\right)=1$. So, $f\left(\frac{\pi}{2}+2\pi n, \frac{\pi}{2}+2\pi n\right) = \frac{\pi}{2}+2\pi n$. This satisfies the constraint $x=y$ and, since $n$ can be arbitrarily large or small, has no absolute maximum or minimum.

Alternately, if we set $x=y$, then $f(x,y)=f(x,x)=x\sin x$. This is easy enough to sketch, and then it is easy enough to see that there are no absolute extrema.

Find all solutions to

The minimum value is $2^{\frac{1}{3}} + 2^{-\frac{2}{3}} =\frac{3}{2}\sqrt[3]{2} =\frac{3}{\sqrt[3]{4}}$ at $\big(\pm 2^{\frac{1}{6}}\,,\, 2^{-\frac{1}{3}}\big)$.

So we are to minimize $f(x,y) = x^2+y^2$ subject to the constraint $g(x,y) = x^2 y -1=0$.

The constraint is not a closed curve, so we need to be a little more careful than average. We can interpret our objective function as the distance from the origin squared. So we're trying to find the point on the curve $y=\frac{1}{x^2}$ that is closest to the origin. The distance from points on that curve to the origin can be arbitrarily large, so the system has no absolute maximum. It does have an absolute minimum, which will also be a local minimum, so it will be a solution to the system of Lagrange equations.

According to the method of Lagrange multipliers, we need to find all solutions to

So the two points to check are $\big(2^{\frac{1}{6}}\,,\,2^{-\frac{1}{3}}\big)$ and $\big(-2^{\frac{1}{6}}\,,\,2^{-\frac{1}{3}}\big)$. For both of these critical points,

This is a straightforward application of the method of Lagrange multipliers, Theorem 16.3.2 in the text.

The maximum and minimum values of $f$ are $\frac{1}{2\sqrt{2}}$ and $-\frac{1}{2\sqrt{2}}$, respectively.

For this problem the objective function is $f(x,y) = xy$ and the constraint function is $g(x,y)=x^2 + 2y^2 - 1$. To apply the method of Lagrange multipliers we start by computing the first order derivatives of these functions.

So, according to the method of Lagrange multipliers, we need to find all solutions to

The method of Lagrange multipliers, Theorem 16.3.2 in the text, gives that the only possible locations of the maximum and minimum of the function $f$ are $\left(\pm\frac{1}{\sqrt{2}},\pm\frac{1}{2}\right)$.

So the maximum and minimum values of $f$ are $\frac{1}{2\sqrt{2}}$ and $-\frac{1}{2\sqrt{2}}$, respectively.

This is a straightforward application of the method of Lagrange multipliers, Theorem 16.3.2 in the text.

min$=1$, max$=\sqrt{2}$.

This is a constrained optimization problem with the objective function being $f(x,y) = x^2 + y^2$ and the constraint function being $g(x,y) =x^4 + y^4 - 1$. By Theorem 16.3.2 in the text, any minimum or maximum $(x,y)$ must obey the Lagrange multiplier equations

So the minimum value of $f$ on $x^4+y^4=1$ is $1$ and the maximum value of $f$ on $x^4+y^4=1$ is $\sqrt{2}$.

When you set your two equations for $\lambda$ equal to one another, you should get something that you can easily plug into the constraint function.

absolute min $\frac{13-8\sqrt2}{3}$, absolute max $\frac53$

Comparing:

We want to minimize $\sqrt{x^2+y^2}$; it's easier to minimize $f(x,y)=x^2+y^2$. The minima will occur at the same point $(x,y)$.

Note the system has no maximum, since we can keep travelling along the parabola to end up arbitrarily far from the origin.

$(\pm1,1/2)$

(It's possible to solve this without Lagrange, but we were asked to use Lagrange to practice the technique.)

We want to minimize $\sqrt{x^2+y^2}$, the distance from the origin to a point $(x,y)$. Note the minimum of that function will occur at the same $(x,y)$-values as the minimum of its square, $x^2+y^2$. Since that's easier to minimize, we use it as our objective function: $f(x,y)=x^2+y^2$.

We only care about coordinates that are actually on the parabola, so our constraint function is $g(x,y)=y+x^2=\frac32$.

Our constraint function is not a closed curve. We can keep travelling along the parabola to end up arbitrarily far from the origin. So there's no global maximum distance, but there is a global minimum distance. The global minimum will also be a local minimum, so it will be a solution to the Lagrange equations.

$f\left(0,\frac32\right)=\frac94$ and $f\left(\pm1,\frac12\right)=\frac{5}{4}$. So, the closest points to the origin on the parabola are the points $(-1,1/2)$ and $(1,1/2)$.

To find extrema over a region, we check critical points and the boundary.

Largest $\frac{\sqrt5}{10-2\sqrt5}$, smallest $\frac{-\sqrt5}{10-2\sqrt5}$

To find extrema over a region, we check CPs and the boundary.
$f(x,y)=xy$, so $f_x=y$ and $f_y=x$. Then the only CP is $(0,0)$.

To check the boundary, we need to know the extreme values of $f(x,y)=xy$ over the ellipse $x^2-2xy+5y^2=1$. It seems tough to do this with plugging in, so we use Lagrange.

All together, we've identified 5 possible locations of extrema.

The largest and smallest of these are $\frac{\sqrt5}{10-2\sqrt5}$ and $\frac{-\sqrt5}{10-2\sqrt5}$, respectively.

You can check your answer from (a) by using a method other than Lagrange multipliers.

(a) (i)

(a) (ii) The warmest point is $(0,10)$ and the coolest point is $(0,-10)$.

(b) (i)

(b) (ii) $(0,0)$ and $(0,-2)$

(c) $(0,0)$

By way of preparation, we have

(a) (i) For this problem the objective function is $T(x,y) = e^y\big(x^2+y^2\big)$ and the constraint function is $g(x,y)=x^2 + y^2 - 100$. According to the method of Lagrange multipliers, Theorem 16.3.2 in the text, we need to find all solutions to

(a) (ii)

So the only possible locations of the maximum and minimum of the function $T$ are $(0,10)$ and $(0,-10)$. To complete this part of the problem, we only have to compute $T$ at those points.

Hence the maximum value of $T(x,y) = e^y\big(x^2+y^2\big)$ on $x^2 + y^2 = 100$ is $100 e^{10}$ at $(0,10)$ and the minimum value is $100 e^{-10}$ at $(0,-10)$.

We remark that, on $x^2+y^2=100$, the objective function $T(x,y) = e^y\big(x^2+y^2\big) = 100 e^y$. So of course the maximum value of $T$ is achieved when $y$ is a maximum, i.e. when $y=10$, and the minimum value of $T$ is achieved when $y$ is a minimum, i.e. when $y=-10$.

(b) (i) By definition, the point $(x,y)$ is a critical point of $T(x,y)$ if and only if the first order partial derivatives at that point are both zero, or at least one does not exist. The first partial derivatives

are well defined everywhere and so the critical points are exactly the point where

(b) (ii) Equation (E1) forces $x=0$. When $x=0$, equation (E2) reduces to

So there are two critical points, namely $(0,0)$ and $(0,-2)$.

(c) Note that $T(x,y) = e^y\big(x^2+y^2\big)\ge 0$ on all of $\bbbr^2$. As $T(x,y)=0$ only at $(0,0)$, it is obvious that $(0,0)$ is the coolest point.

In case you didn't notice that, here is a more conventional solution.

The coolest point on the solid disc $x^2+y^2\le 100$ must either be on the boundary, $x^2+y^2= 100$, of the disc or be in the interior, $x^2+y^2 < 100$, of the disc.

In part (a) (ii) we found that the coolest point on the boundary is $(0,-10)$, where $T=100 e^{-10}$.

If the coolest point is in the interior, it must be a critical point and so must be either $(0,0)$, where $T=0$, or $(0,-2)$, where $T= 4e^{-2}$.

So the coolest point is $(0,0)$.

Since $x \ge 0$ and $y \ge 0$, our constraint function has endpoints $(x,y)=\left(0,400\right)$ and $(x,y)=\left(25,0\right)$. Absolute extrema will occur at these endpoints or at points that solve the system of Lagrange equations.

Min 0; max $75\cdot 2^{10/3}$

Since $x \ge 0$ and $y \ge 0$, our constraint function has endpoints $(x,y)=\left(0,400\right)$ and $(x,y)=\left(25,0\right)$. Absolute extrema will occur at these endpoints or at points that solve the system of Lagrange equations.

Since $g_x$ and $g_y$ are always nonzero, we only have one of our usual three cases.

Now we compare our three points of interest.

The constraint tells you $a+2b=1$. So, your variables are $a$ and $b$.

$4$

The constraint tells us

The triangle formed is a right triangle with area $\frac12bh$. Its base and height are the two intercepts of the line. That is, its base is $\frac1a$, and its height is $\frac1b$. So, the area (which we want to minimize) is

By choosing lines with slopes close to 0, or large negative slopes, we can make triangles with arbitrarily large area. So the absolute minimum will occur somewhere in between at a local minimum value. So we can find the absolute minimum using the method of Lagrange multipliers.

Since $g_a$ and $g_b$ can't be 0, we have only one of our usual three cases.