Lagrange Multipliers

By now, the procedure should be fairly familiar. We set up the partials

$f_x = 2y = 4\lambda x^3 = \lambda g_x$ (equation 1)
$f_y = 2x = 4\lambda y^3 = \lambda g_y$ (equation 2)
$x^4 + y^4 = 2$ (equation 3)

Now, with a little astute observation we can see that if $\lambda = 0$ then both x and y are forced to be 0 and this fails to satisfy our constraint equation. Now, we may safely assume that $\lambda \neq 0$. With a little bit more care we can see that if x or y is zero, it forces either the other variable to be zero as well or $\lambda = 0$ and we know this cannot happen. Now, we can multiply through equation 1 by x and equation 2 by y to obtain

$2yx = 4\lambda x^4$ (equation 1*)
$2xy = 4\lambda y^4$ (equation 2*)

We can set these equations equal to one another
$4\lambda x^4 = 4\lambda y^4 \rightarrow$
$x^4 = y^4 \rightarrow$ (equation 4)

And we can substitute equation 4 into equation 3 and solve for y and then subsequently x.
$y^4 + y^4 = 2y^4 = 2 \rightarrow y = \pm 1$ and thus $x = \pm 1$.

Checking these solutions, we find that the max of f(x, y) = 2 at (1, 1) or (-1, -1). We still need to consult the relevant derivatives and we find f has a saddle point at (0, 0) and that is not the maximum.

We first obtain the relevant partials and set them equal yielding:
$f_x = ycos(xy) = 2\lambda x = \lambda g_x$ (equation 1)
$f_y = xcos(xy) = 2\lambda y = \lambda g_y$ (equation 2)
$x^2 + y^2 = 1 \rightarrow$ (equation 3)
$x = \pm \sqrt{1-y^2}$ (equation 3*)

Here, we notice we have 3 equations and 3 unknowns and substituting equation 3* into equation 1 we can obtain
$ycos(y \sqrt{1-y^2}) = 2\lambda \sqrt{1-y^2} \rightarrow$
$\frac{ycos(y \sqrt{1-y^2})}{2 \sqrt{1-y^2}} = \lambda$ (equation 4)

We will then substitute this into equation equation 2 along with equation 3* to obtain
$\sqrt{1-y^2}cos(\sqrt{1-y^2}y) = 2y\frac{ycos(y \sqrt{1-y^2})}{2 \sqrt{1-y^2}} \rightarrow$
$\sqrt{1-y^2}cos(\sqrt{1-y^2}y) = \frac{y^2cos(y \sqrt{1-y^2})}{\sqrt{1-y^2}} \rightarrow$
$\sqrt{1-y^2}cos(\sqrt{1-y^2}y) – \frac{y^2cos(y \sqrt{1-y^2})}{\sqrt{1-y^2}} = 0\rightarrow$
$cos(y\sqrt{1-y^2})(\sqrt{1-y^2} – \frac{y^2}{\sqrt{1-y^2}}) = 0\rightarrow y = \{0, 1, \frac{1}{\sqrt{2}}\}$
Noticing that we also obtain symmetrical values with substitution of the negative expression and with the same work for the x variable we see our full set of solutions is
$x = \{0, \pm1, \pm\frac{1}{\sqrt{2}}\}, y = \{0, \pm1, \pm\frac{1}{\sqrt{2}}\}$

We then examine the solutions pairwise to see which simultaneously solve equation 3 from above and we obtain the pairs $(0, 1), (0, -1), (1, 0), (-1, 0), (\pm\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (\frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}})$

Evaluating all 8 of these in f(x, y) we find two maxes at $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ and two mins at $(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$.

We also need to examine the partials over the interior to the disc and we find $f_x = ycos(xy) = 0 \rightarrow y = 0$ and similarly, $f_y = 0 \rightarrow x = 0$. If we examine the relevant point we find that it is a saddle point on the interior and therefore cannot be a min or a max.

Now, we need to recall the volume a rectangular prism:
$v(x, y, z) = xyz$ (equation 1)

and the by repeated application of the pythagorean theorem and situation the near vertex on the origin and the far one in the positive octant, we have

$10 = \sqrt{x^2 + y^2 + z^2}$ (equation 2)
We will save ourselves some trouble by replacing equation 2 with the equivalent condition
$100 = x^2 + y^2 + z^2$ (equation 2*)

We now set up the relevant partials
$v_x = yz = 2\lambda x = \lambda g_x$ (equation 3)
$v_y = xz = 2\lambda y = \lambda g_y$ (equation 4)
$v_z = xy = 2\lambda z = \lambda g_z$ (equation 5)
$100 = x^2 + y^2 + z^2$ (equation 2*)

To solve this collection of equations most easily, we need to recall a sneaky fact about boxes: all their dimensions are positive so we can rewrite equations 3, 4 and 5 as:
$xyz = 2\lambda x^2$ (equation 3*)
$xyz = 2\lambda y^2$ (equation 4*)
$xyz = 2\lambda z^2$ (equation 5*)

Setting equation 3* and 4* equal to one another, we obtain
$2\lambda (x^2 – y^2) = 0 \rightarrow x = y$ since we cannot have $\lambda = 0$ in this instance (it forces one of the dimensions to be zero) and we must take the positive solution because our box needs positive dimensions. We can repeat this trick with equations 3* and 5* and obtain that $x = z$ and plugging into equation 2* we find

$100 = 3x^2 \rightarrow \frac{10}{\sqrt{3}} = x$

We note that a cube maximizes a rectangular prism given a fixed long diagonal and the resultant volume in this case is $v(x, x, x) = \frac{1000}{3\sqrt{3}}$.