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an-open-rectanqular-container-is-to-have-a-volume-of-62-5cm-3-find-the-least-possible-surface-area-of-the-material-required-




Question Number 115368 by mathdave last updated on 25/Sep/20
an open rectanqular container is to  have a volume of 62.5cm^3 .find the least  possible surface area of the material  required
$${an}\:{open}\:{rectanqular}\:{container}\:{is}\:{to} \\ $$$${have}\:{a}\:{volume}\:{of}\:\mathrm{62}.\mathrm{5}{cm}^{\mathrm{3}} .{find}\:{the}\:{least} \\ $$$${possible}\:{surface}\:{area}\:{of}\:{the}\:{material} \\ $$$${required} \\ $$
Answered by 1549442205PVT last updated on 25/Sep/20
Denote the sizes of bottom of the  open rectanqular container by x,y;  let z is the heigh of the container.  From the hypothesis we have  xyz=62.5(cm^3 ).Then the surface area   the open rectanqular container is  S=2xz+2yz+xy.Applying Cauchy′s  inequality for three positive numbers  we get  S=2xz+2yz+xy≥3^3 (√(2xz.2yz.xy))  =3^3 (√(4(xyz)^2 ))=3^3 (√(4.62.5^2 ))=75  The equality ocurrs if and only if   { ((xyz=62.5)),((2xz=2yz=xy)) :}⇔x=y=5,z=(5/2)  Thus,the least surface area of the  container equal to 75cm^(2 ) when  x=y=5cm,z=2.5cm
$$\mathrm{Denote}\:\mathrm{the}\:\mathrm{sizes}\:\mathrm{of}\:\mathrm{bottom}\:\mathrm{of}\:\mathrm{the} \\ $$$${open}\:{rectanqular}\:{container}\:\mathrm{by}\:\mathrm{x},\mathrm{y}; \\ $$$$\mathrm{let}\:\mathrm{z}\:\mathrm{is}\:\mathrm{the}\:\mathrm{heigh}\:\mathrm{of}\:\mathrm{the}\:\mathrm{container}. \\ $$$$\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{xyz}=\mathrm{62}.\mathrm{5}\left(\mathrm{cm}^{\mathrm{3}} \right).\mathrm{Then}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{area}\: \\ $$$$\mathrm{the}\:{open}\:{rectanqular}\:{container}\:\mathrm{is} \\ $$$$\mathrm{S}=\mathrm{2xz}+\mathrm{2yz}+\mathrm{xy}.\mathrm{Applying}\:\mathrm{Cauchy}'\mathrm{s} \\ $$$$\mathrm{inequality}\:\mathrm{for}\:\mathrm{three}\:\mathrm{positive}\:\mathrm{numbers} \\ $$$$\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{S}=\mathrm{2xz}+\mathrm{2yz}+\mathrm{xy}\geqslant\mathrm{3}\:^{\mathrm{3}} \sqrt{\mathrm{2xz}.\mathrm{2yz}.\mathrm{xy}} \\ $$$$=\mathrm{3}\:^{\mathrm{3}} \sqrt{\mathrm{4}\left(\mathrm{xyz}\right)^{\mathrm{2}} }=\mathrm{3}\:^{\mathrm{3}} \sqrt{\mathrm{4}.\mathrm{62}.\mathrm{5}^{\mathrm{2}} }=\mathrm{75} \\ $$$$\mathrm{The}\:\mathrm{equality}\:\mathrm{ocurrs}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\begin{cases}{\mathrm{xyz}=\mathrm{62}.\mathrm{5}}\\{\mathrm{2xz}=\mathrm{2yz}=\mathrm{xy}}\end{cases}\Leftrightarrow\mathrm{x}=\mathrm{y}=\mathrm{5},\mathrm{z}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{Thus},\mathrm{the}\:\mathrm{least}\:\mathrm{surface}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{container}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{75cm}^{\mathrm{2}\:} \mathrm{when} \\ $$$$\mathrm{x}=\mathrm{y}=\mathrm{5cm},\mathrm{z}=\mathrm{2}.\mathrm{5cm} \\ $$
Commented by mathdave last updated on 25/Sep/20
gud work
$${gud}\:{work} \\ $$

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