The question I am working on is:
Find the most economical dimensions of a closed rectangular box of volume 8 cubic units if the cost of the material per square unit for (i) the top and bottom is 8, (ii) the front and back is 6 and (iii) the other two sides is 2.
I decided to use Lagrange Multipliers to solve this.
The equations I set up to work with were:
$xyz = 8$
$C = 8(2xy) + 6(2xz) + 2(2yz) = 16xy + 12xz + 4yz$
Taking partial derivatives I have:
$\nabla C = \lambda\nabla V = <16y + 21z, 16x + 4z, 12x + 4y>$
$\lambda xyz = x(16y + 12z) = y(16x + 4z) = z(12x+4y)$
Solving for x and z, I get: $ x = y/3$ and $z = 4y/3$
Plugging into $xyz = 8$, $y = 2.621$. I then plugged that into the x and z equations to get $.874$ and $3.494$ respectively. None of these values are correct according to the WebWork.
Where did I go wrong? Thanks in advance!
This can be done without lagrange.
$V=8, V=xyz$
Top and Bottom: $xy=8$ sides: $xz=2$ front and back: $yz=6$
Want to minimize $C=8(2xy)+2(2xz)+6(2yz)$
$C= 16xy+4xz+12yz$
now, we have $xyz=8$ so $$z=\frac{8}{xy}$$
subbing into C gives $$C= 16xy+4x\frac{8}{xy}+12y\frac{8}{xy}$$
$$C= 16xy+\frac{32}{y}+\frac{96}{x}$$
Now computing partials and setting to zero,
$$C_x= 16y-\frac{96}{x^2}=0$$
$\rightarrow 16x^2y=96$
$$C_y= 16x-\frac{32}{y^2}=0$$
$16y^2x=32$
Multiplying by y and x respectively,
$16x^2y^2=32x=96y$
$\rightarrow x=3y$
using $C_x=0$ to solve $$16(3y)-\frac{32}{y^2}=0 \rightarrow y= (2/3)^{1/3}$$
from which x easily follows (x=3y) and knowing both x and y you can solve for z easily as well, in a few ways but may be easies to plug in x and y into 8=xyz.