Hello, ikickyouindanuts!
Find the dimensions of an open-top rectangular box with a volume of 32 ft³,
using the least amount of material. You need to be familiar with partial derivatives . . . Code:
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/ | / |H
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L
The volume is 32 ft³: .$\displaystyle LWH \:=\:32\quad\Rightarrow\quad H \:=\:\frac{32}{LW}\;\;{\color{blue}1]}$
The base has area: .$\displaystyle LW$
The front and back has area: .$\displaystyle 2LH$
The sides have area: .$\displaystyle 2WH$
The total area is: .$\displaystyle A \:=\:LW + 2LH + 2WH\;\;{\color{blue}[2]}$
Substitute [1] into [2]: .$\displaystyle A \;=\;LW + 2L\left(\frac{32}{LW}\right) + 2W\left(\frac{32}{LW}\right)$
. . and we have: .$\displaystyle A \;=\;LW + 64W^{-1} + 64L^{-1}$
Take partial dereivatives and equate to zero:
. . $\displaystyle \begin{array}{cccccccccc}\frac{\partial A}{\partial L} &=& W - 64L^{-2} &=& 0 & \Rightarrow & L^2W &=& 64 & {\color{blue}[3]}\\ \frac{\partial A}{\partial W} &=& L - 64W^{-2} &=&0 & \Rightarrow & LW^2 &=& 64 & {\color{blue}[4]}\end{array}$
Equate [3] and [4]: .$\displaystyle L^2W \:=\:LW^2\quad\Rightarrow\quad L \:=\:W$
Substitute into [3]: .$\displaystyle L^3 \:=\:64\quad\Rightarrow\quad L \:=\:4\quad\Rightarrow\quad W \:=\:4$
Substitute into [1]: .$\displaystyle H \:=\:\frac{32}{4\cdot4} \;=\;2$
. . Therefore: .$\displaystyle (L,W,H) \;=\;(4,4,2)$