1. ## Larange Multipliers Problem

I am having a bit of trouble setting up a problem:

(1) Find the volume of the largest regtangular box with edges parallel to the axes that can be inscribed in the ellipsoid: 9x^2 + 36y^2 + 4z^2 = 36

I thought it would be simply:

f(x,y,z) = xyz && g(x,y,z) = 9x^2 + 36y^2 + 4z^2 = 36

However, this seems to give me the wrong solution.

Any help would be appreciated.

2. Originally Posted by machi4velli
I am having a bit of trouble setting this problem up:

Find the volume of the largest regtangular box with edges parallel to the axes that can be inscribed in the ellipsoid: 9x^2 + 36y^2 + 4z^2 = 36

Any help would be appreciated.
Let the box have width $w$, depth $d$, and height $h$.

Then:

$
(h/2)^2 \le 36/4
$

Also:

$
(w/2)^2 \le \frac{36-4(h/2)^2}{9}
$

and:

$
(d/2)^2 = \frac{36-4(h/2)^2-9(w/2)^2}{36}
$
.

So our problem becomes:

Maximise $f(h,w,d) = h\times w \times d$ subject to:

$
(h/2)^2 \le 9
$

$
(w/2)^2 \le \frac{36-4(h/2)^2}{9}
$

$
(d/2)^2 = \frac{36-4(h/2)^2-9(w/2)^2}{36}
$
.

RonL

3. Originally Posted by machi4velli
I am having a bit of trouble setting up a problem:

(1) Find the volume of the largest regtangular box with edges parallel to the axes that can be inscribed in the ellipsoid: 9x^2 + 36y^2 + 4z^2 = 36

I thought it would be simply:

f(x,y,z) = xyz && g(x,y,z) = 9x^2 + 36y^2 + 4z^2 = 36

However, this seems to give me the wrong solution.

Any help would be appreciated.
I agree with your setup except that you need to multiply xyz by a constant to get the volume.

In two dimensions with an ellipse,

max xy subject to 9x^2 + 36y^2 = 36

would produce a point (x,y) on the ellipse in the upper right quadrant. The inscribed rectangle would have corners (x,y), (x,-y), (-x,y) and (-x,-y). The area of the rectangle would be 4xy.

In three dimensions, the constant to multiply xyz by is 8.

4. Hello, machi4velli!

(1) Find the volume of the largest regtangular box with edges parallel to the axes
that can be inscribed in the ellipsoid: . $9x^2+36y^2+4z^2\:=\:36$
This is the way I was taught to use Lagrange Multipliers.

The length of the box is $2x$, the width $2y$, the height $2z$.
. . The volume of the box is: . $(2x)(2y)(2z) \,=\,8xyz$.

We have: . $V(x,y,z)\:=\:8xyz$ .and . $g(x) \:=\:9x^2+36y^2+4z^2 -36$

. . Then: . $f(x,y,z,\lambda) \;=\;8xyz + \lambda(9x^2 + 36y^2 + 4z^2 - 36)$

Set the four partial derivatives equal to zero, and solve.

[1] . $f_x \:=\:8yz + 18x\lambda\:=\:0$

[2] . $f_y \:=\:8xz + 72y\lambda\:=\:0$

[3] . $f_z \:=\:8xy + 8z\lambda \:=\:0$

[4] . $f_{\lambda} \:=\:9x^2+36y^2 + 4z^2 - 36 \:=\:0$

From [1], we have: . $\lambda \:=\:-\frac{8yz}{18x}\:=\:-\frac{4y^2}{9x}$ .[5]

From [2], we have: . $\lambda \:=\:-\frac{8xz}{72y}\:=\:-\frac{xz}{9y}$ .[6]

From [3], we have: . $\lambda\:=\:-\frac{8xy}{8z}\:=\:-\frac{xy}{z}$ .[7]

Equate [5] and [6]: . $-\frac{4yz}{9x}\:=\:-\frac{xz}{9y}\quad\Rightarrow\quad x^2\,=\,4y^2$ .[8]

Equate [6] and [7]: . $-\frac{xz}{9y} \:=\:-\frac{xy}{9z}\quad\Rightarrow\quad z^2 \,=\,9y^2$ .[9]

Substitute [8] and [9] into [4]: . $9(4y^2) + 36y^2 + 4(9y^2) - 36 \:=\:0$

. . and we get: . $108y^2\:=\:36\quad\Rightarrow\quad y^2\:=\:\frac{1}{3}$

Substitute into [8]: . $x^2\:=\:4\left(\frac{1}{3}\right) \:=\:\frac{4}{3}$
Substitute into [9]: . $z^2 \:=\:9\left(\frac{1}{3}\right)\:=\:3$

. . Hence: . $x\:=\:\frac{2\sqrt{3}}{3},\;y \:=\:\frac{\sqrt{3}}{3},\;z\:=\:\sqrt{3}$

Therefore, the maximum volume is: . $8\left(\frac{2\sqrt{3}}{3}\right)\left(\frac{\sqrt {3}}{3}\right)\left(\sqrt{3}\right) \:=\:\frac{16\sqrt{3}}{3}$ unitsł

5. heheh, the 8 threw me off, i was getting 2sqrt3/3, thanks a lot guys