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Question Number 107658 by john santu last updated on 12/Aug/20

$$\divideontimes \mathcal{JS}\divideontimes$$$$Ametalbox\left(withouttop\right)istobe$$$$contructedfromasquaresheet$$$$ofmetalthatis10dmonthe$$$$sidebyfirstcuttingthesquare$$$$piecesofthesamesizefromthe$$$$cornersofthesheetandfolding$$$$upsides.whatsizesquares$$$$shouldbecutinorderto$$$$maximizethevolumeofthebox.$$

Answered by som(math1967) last updated on 12/Aug/20

$$\mathrm{let}\mathrm{side}\mathrm{of}\mathrm{square}(\mathrm{cut}\mathrm{from}$$$$\mathrm{corner})\mathrm{xdm}$$$$\therefore \mathrm{volume}\mathrm{V}={(10-2\mathrm{x})}^2\times \mathrm{x}$$$$\Rightarrow \mathrm{V}=100\mathrm{x}-{40\mathrm{x}}^2+{4\mathrm{x}}^3$$$$\frac{\mathrm{dV}}{\mathrm{dx}}=100-80\mathrm{x}+{12\mathrm{x}}^2$$$$\frac{{\mathrm{d}}^2\mathrm{V}}{{\mathrm{dx}}^2}=24\mathrm{x}-80$$$$\mathrm{For}{\mathrm{V}}_{\max }/{\mathrm{V}}_{\min }\frac{\mathrm{dV}}{\mathrm{dx}}=0$$$$\therefore {12\mathrm{x}}^2-80\mathrm{x}+100=0$$$${3\mathrm{x}}^2-20\mathrm{x}+25=0$$$$(\mathrm{x}-5)(3\mathrm{x}-5)=0$$$$\mathrm{x}=5,\frac{5}{3}$$$$\mathrm{now}{\left(\frac{{\mathrm{d}}^2\mathrm{V}}{{\mathrm{dx}}^2}\right)}_5=40>0$$$${\left(\frac{{\mathrm{d}}^2\mathrm{V}}{{\mathrm{dx}}^2}\right)}_{\frac{5}{3}}=-40<0$$$$\therefore \mathrm{for}\frac{5}{3}{\mathrm{V}}_{\max }\mathrm{obtain}$$$$\therefore \mathrm{side}\mathrm{of}\mathrm{square}=\frac{5}{3}\mathrm{dm}$$

Commented by som(math1967) last updated on 12/Aug/20

Commented by bemath last updated on 12/Aug/20

$$sirhowdoyoumakeit?$$

Commented by som(math1967) last updated on 12/Aug/20

$$\mathrm{using}\mathrm{insert}\mathrm{drawing}\left(\mathrm{Tinkutara}\right)$$

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