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

⋇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 folding up sides. what size squares should be cut in order to maximize the volume of the box.

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

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

$$ \\ $$$$ \\ $$

Commented by bemath last updated on 12/Aug/20

sir how do you make it ?

$${sir}\:{how}\:{do}\:{you}\:{make}\:{it}\:?\: \\ $$

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

using insert drawing(Tinkutara)

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

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