Optimization with Partials

[heartshapes]

Basically, a rectangular box needs to hold 512cm[sup:kletuwg0]3[/sup:kletuwg0]. The sides of the box cost $0.01/cm[sup:kletuwg0]2[/sup:kletuwg0] and the top and bottom cost $0.02/cm[sup:kletuwg0]2[/sup:kletuwg0]. Find the dimensions that minimize the total cost of materials used.

This is what I have done so far.
\(\displaystyle C=.02(2lw)+.01(2lh+2wh)\\\)
\(\displaystyle 512=lwh\\\)

Solved for w.
\(\displaystyle w=\frac{512}{lh}\\\)
\(\displaystyle C=.02(2l(\frac{512}{lh}))+.01(2lh+2\frac{512}{lh}h)\\\)
\(\displaystyle C=.04(\frac{512}{h})+.02lh+.02(\frac{512}{l})\\\)
\(\displaystyle C=\frac{20.48}{h}+.02lh+\frac{10.24}l{}\\)

I found the partials
\(\displaystyle C_{h}=\frac{-20.48}{h^{2}}+.02l C\)
\(\displaystyle C_{l}=.02h-\frac{10.24}{l^{2}}\)
Then set them equal to zero.

\(\displaystyle \frac{-20.48}{h^{2}}+.02l C =0\)
\(\displaystyle .02h-\frac{10.24}{l^{2}} =0\)

Then, I solved for h.
\(\displaystyle h=\frac{10.24}{.02l^{2}}\)

Plugged h in \(\displaystyle C_{h}\)
\(\displaystyle 0=\frac{-20.48}{\frac{10.24}{.02l^{2}}^{2}}+.02l\)

This is where I get confused. I have tried to simplify it so many times and I end up with funny numbers. Any help or direction would be greatly appreciated.

 

[soroban]

Hello, heartshapes!

First of all, I'd drop the decimal points . . .

A rectangular box needs to hold 512cm[sup:mjpkd6zx]3[/sup:mjpkd6zx].
The sides of the box cost 1/cm[sup:mjpkd6zx]2[/sup:mjpkd6zx] and the top and bottom cost 2/cm[sup:mjpkd6zx]2[/sup:mjpkd6zx].

Find the dimensions that minimize the total cost of materials used.

I began the same as you . . .

\(\displaystyle C \;=\;2(2LW) + 1(2LH + 2WH) \;=\;4LW + 2LH + 2WH\)

\(\displaystyle LWH \:=\:512 \quad\Rightarrow\quad W \:=\:\frac{512}{LH}\;\;[1]\)

\(\displaystyle \text{Substitute: }\;C \;=\;4L\left(\frac{512}{LH}\right) + 4WH + 2H\left(\frac{512}{LH}\right)\)

. . \(\displaystyle \text{and we have: }\;C \;=\;2048H^{-1} + 2LH + 1024L^{-1}\)


Equate partial derivatives to zero:

. . \(\displaystyle C_H \;=\;-2048H^{-2} + 2L \:=\:0 \quad\Rightarrow\quad LH^2 \:=\:1024\;\;[2]\)

. . \(\displaystyle C_L \;=\;2H - 1024L^{-2} \:=\:0 \quad\Rightarrow\quad L^2H \:=\:512\;\;[3]\)


\(\displaystyle \text{Divide [2] by [3]: }\;\frac{LH^2}{L^2\!H} \:=\:\frac{1024}{512} \quad\Rightarrow\quad \frac{H}{L} \:=\:2 \quad\Rightarrow\quad H \:=\:2L\;\;[4]\)

\(\displaystyle \text{Substitute into [3]: }\;L^2(2L) \:=\:512 \quad\Rightarrow\quad L^3 \:=\:256 \quad\Rightarrow\quad\boxed{ L \:=\:4\sqrt[3]{4}}\)

\(\displaystyle \text{Substitute into [4]: }\;H \:=\:2L \quad\Rightarrow\quad\boxed{ H\;=\;8\sqrt[3]{4}}\)

\(\displaystyle \text{Substitute into [1]: }\;W \:=\:\frac{512}{LH} \:=\:\frac{512}{(4\sqrt[3]{4})(8\sqrt[3]{4})} \quad\Rightarrow\quad \boxed{W \:=\:4\sqrt[3]{4}}\)


\(\displaystyle \text{Therefore: }\;L \times W \times H \;=\;4\sqrt[3]{4} \:\times \:4\sqrt[3]{4} \:\times \:8\sqrt[3]{4}\)

 

[heartshapes]

Thank you! I actually forgot to type the decimals in the question. Oops.
When I did it I got 1.6 x 3.2 x 100. Atleast I got an answer though. I am going to look over it later. THANK YOU!