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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
   ⋇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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