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An-open-box-of-area-486cm-2-If-the-length-is-twice-the-breadth-Find-the-maximum-volume-of-the-box-hence-Show-the-volume-is-maximum-

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Question Number 14543 by chux last updated on 02/Jun/17
An open box of area 486cm^2 .If the length is twice the breadth.Find the maximum volume of the box. hence,Show the volume is maximum.
$$\mathrm{An}\:\mathrm{open}\:\mathrm{box}\:\mathrm{of}\:\mathrm{area}\:\mathrm{486cm}^{\mathrm{2}} .\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{length}\:\mathrm{is}\:\mathrm{twice}\:\mathrm{the}\:\mathrm{breadth}.\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{box}. \\ $$$$\mathrm{hence},\mathrm{Show}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{is}\:\mathrm{maximum}. \\ $$
Commented by chux last updated on 02/Jun/17
please help
$$\mathrm{please}\:\mathrm{help} \\ $$
Answered by sandy_suhendra last updated on 22/Jun/17
let the breadth=x the length=2x and the height=h the area of the box=2x.x+(2x+x+2x+x)h 486 = 2x^2 +6hx 6hx=486−2x^2 h=((486−2x^2 )/(6x))=((243−x^2 )/(3x)) the volume=V=2x.x.h=2x^2 (((243−x^2 )/(3x)))=162x−(2/3)x^3 (dV/dx)=162−2x^2 =0 ⇒ x=9 V max=162×9−(2/3)×9^3 = 972 cm^3
$$\mathrm{let}\:\mathrm{the}\:\mathrm{breadth}=\mathrm{x} \\ $$$$\mathrm{the}\:\mathrm{length}=\mathrm{2x} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{height}=\mathrm{h} \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{box}=\mathrm{2x}.\mathrm{x}+\left(\mathrm{2x}+\mathrm{x}+\mathrm{2x}+\mathrm{x}\right)\mathrm{h} \\ $$$$\mathrm{486}\:=\:\mathrm{2x}^{\mathrm{2}} +\mathrm{6hx} \\ $$$$\mathrm{6hx}=\mathrm{486}−\mathrm{2x}^{\mathrm{2}} \\ $$$$\mathrm{h}=\frac{\mathrm{486}−\mathrm{2x}^{\mathrm{2}} }{\mathrm{6x}}=\frac{\mathrm{243}−\mathrm{x}^{\mathrm{2}} }{\mathrm{3x}} \\ $$$$\mathrm{the}\:\mathrm{volume}=\mathrm{V}=\mathrm{2x}.\mathrm{x}.\mathrm{h}=\mathrm{2x}^{\mathrm{2}} \left(\frac{\mathrm{243}−\mathrm{x}^{\mathrm{2}} }{\mathrm{3x}}\right)=\mathrm{162x}−\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} \:\:\:\:\: \\ $$$$\frac{\mathrm{dV}}{\mathrm{dx}}=\mathrm{162}−\mathrm{2x}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:\mathrm{x}=\mathrm{9} \\ $$$$\mathrm{V}\:\mathrm{max}=\mathrm{162}×\mathrm{9}−\frac{\mathrm{2}}{\mathrm{3}}×\mathrm{9}^{\mathrm{3}} \:=\:\mathrm{972}\:\mathrm{cm}^{\mathrm{3}} \\ $$
Answered by Tinkutara last updated on 02/Jun/17
l = 2b Let height = h Since box is open, area will exclude the topmost surface area. Hence, 2h(l + b) + lb = 486 6bh + 2b^2 = 486 b^2 + 3bh = 243 3bh = 243 − b^2 h = ((243 − b^2 )/(3b)) V = lbh = 2b^2 h = 2b(((243 − b^2 )/3)) = ((486b − 2b^3 )/3) (dV/db) = (1/3)(486 − 6b^2 ) = 0 gives b = 9 (d^2 V/db^2 ) = (1/3)(−12b) which is always negative since b > 0. Volume is maximum at b = 9 cm V_(max) = ((486×9 − 2×729)/3) = 486×3 − 2×243 = 972 cm^3
$${l}\:=\:\mathrm{2}{b} \\ $$$$\mathrm{Let}\:\mathrm{height}\:=\:{h} \\ $$$$\mathrm{Since}\:\mathrm{box}\:\mathrm{is}\:\mathrm{open},\:\mathrm{area}\:\mathrm{will}\:\mathrm{exclude} \\ $$$$\mathrm{the}\:\mathrm{topmost}\:\mathrm{surface}\:\mathrm{area}.\:\mathrm{Hence}, \\ $$$$\mathrm{2}{h}\left({l}\:+\:{b}\right)\:+\:{lb}\:=\:\mathrm{486} \\ $$$$\mathrm{6}{bh}\:+\:\mathrm{2}{b}^{\mathrm{2}} \:=\:\mathrm{486} \\ $$$${b}^{\mathrm{2}} \:+\:\mathrm{3}{bh}\:=\:\mathrm{243} \\ $$$$\mathrm{3}{bh}\:=\:\mathrm{243}\:−\:{b}^{\mathrm{2}} \\ $$$${h}\:=\:\frac{\mathrm{243}\:−\:{b}^{\mathrm{2}} }{\mathrm{3}{b}} \\ $$$${V}\:=\:{lbh}\:=\:\mathrm{2}{b}^{\mathrm{2}} {h}\:=\:\mathrm{2}{b}\left(\frac{\mathrm{243}\:−\:{b}^{\mathrm{2}} }{\mathrm{3}}\right) \\ $$$$=\:\frac{\mathrm{486}{b}\:−\:\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{3}} \\ $$$$\frac{{dV}}{{db}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{486}\:−\:\mathrm{6}{b}^{\mathrm{2}} \right)\:=\:\mathrm{0}\:\mathrm{gives}\:{b}\:=\:\mathrm{9} \\ $$$$\frac{{d}^{\mathrm{2}} {V}}{{db}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(−\mathrm{12}{b}\right)\:\mathrm{which}\:\mathrm{is}\:\mathrm{always} \\ $$$$\mathrm{negative}\:\mathrm{since}\:{b}\:>\:\mathrm{0}. \\ $$$$\mathrm{Volume}\:\mathrm{is}\:\mathrm{maximum}\:\mathrm{at}\:{b}\:=\:\mathrm{9}\:\mathrm{cm} \\ $$$${V}_{\mathrm{max}} \:=\:\frac{\mathrm{486}×\mathrm{9}\:−\:\mathrm{2}×\mathrm{729}}{\mathrm{3}} \\ $$$$=\:\mathrm{486}×\mathrm{3}\:−\:\mathrm{2}×\mathrm{243}\:=\:\mathrm{972}\:\mathrm{cm}^{\mathrm{3}} \\ $$

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