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Question Number 144532 by EDWIN88 last updated on 26/Jun/21

$$\mathrm{A}\mathrm{rectangular}\mathrm{box},\mathrm{open}\mathrm{at}\mathrm{the}$$$$\mathrm{top}\mathrm{is}\mathrm{to}\mathrm{have}\mathrm{a}\mathrm{volume}\mathrm{of}32\mathrm{cube}\mathrm{feet}$$$$\mathrm{What}\mathrm{must}\mathrm{be}\mathrm{the}\mathrm{dimensions}$$$$\mathrm{so}\mathrm{that}\mathrm{the}\mathrm{total}\mathrm{surface}\mathrm{is}\mathrm{a}\mathrm{minimum}?$$

Answered by liberty last updated on 26/Jun/21

$$\mathrm{volume}\mathrm{of}\mathrm{box}=\mathrm{V}=\mathrm{xyz}=32$$$$\mathrm{and}\mathrm{z}=\frac{32}{\mathrm{xy}}$$$$\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{box}=\mathrm{S}=\mathrm{xy}+2\mathrm{yz}+2\mathrm{xz}$$$$\{\begin{array}{l}\frac{\partial \mathrm{S}}{\partial \mathrm{x}}=\mathrm{y}-\frac{64}{{\mathrm{x}}^2}=0\mathrm{when}{\mathrm{x}}^2\mathrm{y}=64\\ \frac{\partial \mathrm{S}}{\partial \mathrm{y}}=\mathrm{x}-\frac{64}{{\mathrm{y}}^2}=0\mathrm{when}{\mathrm{xy}}^2=64\end{array}$$$$\mathrm{so}\mathrm{we}\mathrm{get}\mathrm{y}=\mathrm{x}\mathrm{and}{\mathrm{x}}^3=64\mathrm{or}$$$$\mathrm{x}=\mathrm{y}=4\mathrm{and}\mathrm{z}=2$$$$\mathrm{for}\mathrm{x}=\mathrm{y}=4,\mathrm{\Delta }={\mathrm{S}}_{\mathrm{xx}}{\mathrm{S}}_{\mathrm{yy}}-{\mathrm{S}}_{\mathrm{xy}}^2$$$$\mathrm{\Delta }=\left(\frac{128}{{\mathrm{x}}^3}\right)\left(\frac{128}{{\mathrm{y}}^3}\right)-1>0$$$$\mathrm{and}{\mathrm{S}}_{\mathrm{xx}}=\frac{128}{{\mathrm{x}}^3}>0\mathrm{hence}\mathrm{it}\mathrm{follows}$$$$\mathrm{that}\mathrm{the}\mathrm{dimensions}4\mathrm{feet}\times 4\mathrm{feet}\times 2\mathrm{feet}$$$$\mathrm{give}\mathrm{the}\mathrm{minimum}\mathrm{surface}$$

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