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Of-all-rectangular-boxes-without-a-lid-and-having-a-given-surface-area-Find-the-one-with-maximum-volume-

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Question Number 6135 by sanusihammed last updated on 15/Jun/16
Of all rectangular boxes without a lid and having a given surface area . Find the one with maximum volume.
$${Of}\:{all}\:{rectangular}\:{boxes}\:{without}\:{a}\:{lid}\:{and}\:{having}\:{a}\:{given}\: \\ $$$${surface}\:{area}\:.\:{Find}\:{the}\:{one}\:{with}\:{maximum}\:{volume}. \\ $$
Commented by FilupSmith last updated on 15/Jun/16
Edge lengths a, b, c Max volume when a=b=c V=abc=a^3 =b^3 =c^3
$$\mathrm{Edge}\:\mathrm{lengths}\:{a},\:{b},\:{c} \\ $$$$\mathrm{Max}\:\mathrm{volume}\:\mathrm{when}\:{a}={b}={c} \\ $$$${V}={abc}={a}^{\mathrm{3}} ={b}^{\mathrm{3}} ={c}^{\mathrm{3}} \\ $$
Answered by Rasheed Soomro last updated on 16/Jun/16
Let a,b and c are dimenntions of rectangular box. The surface area =2(ab+bc+ca) Let the lid has a and c dimentions The area of lid surface=ac A=Without lid surface area=2(ab+bc+ca)−ac =2ab+2bc+ca Let there is a cubic box (without a lid ) of dimention d having the surface area equal to that of above box. A=Without lid surface area=d^2 +2d^2 +d^2 =5d^2 ∴ 2ab+2bc+ca=5d^2 d=(√((2ab+2bc+ca)/5)) Now let′s compare the volumes of above boxes Volume of rectangular box=abc Volume of cubic box=d^3 =(((2ab+2bc+ca)/5))^(3/2) (((2ab+2bc+ca)/5))^(3/2) ? abc ∀ a,b,c>0 Assumption: (((2ab+2bc+ca)/5))^(3/2) ≥ abc ∀ a,b,c>0 Can′t continue
$${Let}\:\boldsymbol{{a}},\boldsymbol{{b}}\:{and}\:\boldsymbol{{c}}\:{are}\:{dimenntions}\:{of}\: \\ $$$${rectangular}\:{box}. \\ $$$${The}\:{surface}\:{area}\:=\mathrm{2}\left(\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ca}}\right) \\ $$$${Let}\:{the}\:{lid}\:{has}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{c}}\:{dimentions} \\ $$$${The}\:{area}\:{of}\:{lid}\:{surface}=\boldsymbol{{ac}} \\ $$$$\boldsymbol{{A}}={Without}\:{lid}\:{surface}\:{area}=\mathrm{2}\left(\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ca}}\right)−\boldsymbol{{ac}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\boldsymbol{{ab}}+\mathrm{2}\boldsymbol{{bc}}+\boldsymbol{{ca}} \\ $$$$ \\ $$$${Let}\:{there}\:{is}\:{a}\:{cubic}\:{box}\:\left({without}\:{a}\:{lid}\:\right)\:{of}\:{dimention}\:\boldsymbol{{d}} \\ $$$${having}\:{the}\:{surface}\:{area}\:{equal}\:{to}\:{that}\:{of}\:{above}\:{box}. \\ $$$$\boldsymbol{{A}}={Without}\:{lid}\:{surface}\:{area}=\boldsymbol{{d}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{d}}^{\mathrm{2}} +\boldsymbol{{d}}^{\mathrm{2}} =\mathrm{5}\boldsymbol{{d}}^{\mathrm{2}} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{{ab}}+\mathrm{2}\boldsymbol{{bc}}+\boldsymbol{{ca}}=\mathrm{5}\boldsymbol{{d}}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{d}}=\sqrt{\frac{\mathrm{2}\boldsymbol{{ab}}+\mathrm{2}\boldsymbol{{bc}}+\boldsymbol{{ca}}}{\mathrm{5}}} \\ $$$$ \\ $$$${Now}\:{let}'{s}\:{compare}\:{the}\:{volumes}\:{of}\:{above}\:{boxes} \\ $$$${Volume}\:{of}\:{rectangular}\:{box}=\boldsymbol{{abc}} \\ $$$${Volume}\:{of}\:{cubic}\:\:{box}=\boldsymbol{{d}}^{\mathrm{3}} =\left(\frac{\mathrm{2}\boldsymbol{{ab}}+\mathrm{2}\boldsymbol{{bc}}+\boldsymbol{{ca}}}{\mathrm{5}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\left(\frac{\mathrm{2}\boldsymbol{{ab}}+\mathrm{2}\boldsymbol{{bc}}+\boldsymbol{{ca}}}{\mathrm{5}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} ?\:\:\boldsymbol{{abc}}\:\:\:\forall\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}>\mathrm{0} \\ $$$${Assumption}: \\ $$$$\left(\frac{\mathrm{2}\boldsymbol{{ab}}+\mathrm{2}\boldsymbol{{bc}}+\boldsymbol{{ca}}}{\mathrm{5}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \geqslant\:\:\boldsymbol{{abc}}\:\:\:\forall\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}>\mathrm{0} \\ $$$${Can}'{t}\:{continue} \\ $$

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