JS-A-metal-box-without-top-is-tobe-contructed-from-a-square-sheet-of-metal-that-is-10-dm-on-the-side-by-first-cutting-the-square-pieces-of-the-same-size-from-the-corners-of-the-sheet-and-foldin

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

$$\:\:\:\divideontimes\mathcal{JS}\divideontimes \\ $$$${A}\:{metal}\:{box}\:\left({without}\:{top}\right)\:{is}\:{tobe} \\ $$$${contructed}\:{from}\:{a}\:{square}\:{sheet} \\ $$$${of}\:{metal}\:{that}\:{is}\:\mathrm{10}\:{dm}\:{on}\:{the} \\ $$$${side}\:{by}\:{first}\:{cutting}\:{the}\:{square} \\ $$$${pieces}\:{of}\:{the}\:{same}\:{size}\:{from}\:{the} \\ $$$${corners}\:{of}\:{the}\:{sheet}\:{and}\:{folding} \\ $$$${up}\:{sides}.\:{what}\:{size}\:{squares} \\ $$$${should}\:{be}\:{cut}\:{in}\:{order}\:{to} \\ $$$${maximize}\:{the}\:{volume}\:{of}\:{the}\:{box}. \\ $$

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

let side of square(cut from corner) xdm ∴volume V=(10−2x)^2 ×x ⇒V=100x−40x^2 +4x^3 (dV/dx)=100−80x+12x^2 (d^2 V/dx^2 )=24x−80 For V_(max) /V_(min) (dV/dx)=0 ∴12x^2 −80x+100=0 3x^2 −20x+25=0 (x−5)(3x−5)=0 x=5,(5/3) now ((d^2 V/dx^2 ))_5 =40>0 ((d^2 V/dx^2 ))_(5/3) =−40<0 ∴ for (5/3) V_(max) obtain ∴side of square=(5/3)dm

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

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