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 $2$.

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

Not much calculation is necessary.

There are none

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 $\big(\pm 2^{\frac{1}{6}}\,,\, 2^{-\frac{1}{3}}\big)$.

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.

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

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

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 $\frac53$

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.

$(\pm1,1/2)$

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

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

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) $(0,0)$

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 $75\cdot 2^{10/3}$

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

$4$

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through the point $(1,2)$ if and only if $\frac{1}{a^2}+\frac{4}{b^2}=1$.

$a=b=\sqrt{5}$

You may choose your coordinate system so the cylinder is oriented vertically along the $z$-axis. Then you can write the volume of the cylinder as a function of two variables.

$\text{radius}=\sqrt{\frac{2}{3}}$ and $\text{height}=\frac{2}{\sqrt{3}}$.

The volume is your constraint function.

$3 \times 6 \times 4$

See the solution.

The surface $z=f(x,y)$ is similar to the quadric surface from Example 14.3.2.

Absolute minimum is 0, achieved at $(0,1)$. There is no absolute maximum.

No great amount of computation is needed

There are none.

Although $f(x,y)$ is unbounded, and $x=y$ is not a closed curve, there are indeed absolute extrema of $f(x,y)$ subject to $x=y$. To find them, remember last semester's methods for finding extrema of functions of a single variable.