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Question Number 100341 by peter frank last updated on 26/Jun/20

An open box with a square base is to be made out of a given quantity of a cardboard of area c^2 square units.show the maximum volume of the box (c^2 /(6(√3))) cubic units

$$\mathrm{An}\:\mathrm{open}\:\mathrm{box}\:\mathrm{with}\:\mathrm{a}\:\mathrm{square} \\ $$$$\mathrm{base}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{made}\:\mathrm{out} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{given}\:\mathrm{quantity}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{cardboard}\:\mathrm{of}\:\mathrm{area}\:\mathrm{c}^{\mathrm{2}} \\ $$$$\mathrm{square}\:\mathrm{units}.\mathrm{show}\:\mathrm{the} \\ $$$$\mathrm{maximum}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{box}\:\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{6}\sqrt{\mathrm{3}}}\:\:\mathrm{cubic}\:\mathrm{units} \\ $$$$ \\ $$

Commented by PRITHWISH SEN 2 last updated on 26/Jun/20

Let the area of the floor = x^2 the area of the remaining 4 sides = c^2 −x^2 ∴ the height of tbe box = ((c^2 −x^2 )/(4x)) ∴ The vol. V(x)=(1/4)(c^2 −x^2 )x V^′ (x)=(1/4)(c^2 −3x^2 )=0⇒x^2 =(c^2 /3) V^(′′) (x)=−(3/2)x<0 ∀x ∴V_(max.) = (1/4)(c^2 −(c^2 /3))(c/(√3)) = (c^3 /(6(√3))) proved.

$$\mathrm{Let}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{floor}\:=\:\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{4}\:\mathrm{sides}\:=\:\mathrm{c}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \\ $$$$\therefore\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{tbe}\:\mathrm{box}\:=\:\frac{\mathrm{c}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }{\mathrm{4x}} \\ $$$$\therefore\:\mathrm{The}\:\mathrm{vol}.\:\mathrm{V}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{c}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \right)\mathrm{x} \\ $$$$\:\:\:\:\:\:\:\mathrm{V}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{c}^{\mathrm{2}} −\mathrm{3x}^{\mathrm{2}} \right)=\mathrm{0}\Rightarrow\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{V}^{''} \left(\mathrm{x}\right)=−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}<\mathrm{0}\:\:\:\forall\mathrm{x} \\ $$$$\therefore\boldsymbol{\mathrm{V}}_{\boldsymbol{\mathrm{max}}.} =\:\frac{\mathrm{1}}{\mathrm{4}}\left(\boldsymbol{\mathrm{c}}^{\mathrm{2}} −\frac{\boldsymbol{\mathrm{c}}^{\mathrm{2}} }{\mathrm{3}}\right)\frac{\boldsymbol{\mathrm{c}}}{\sqrt{\mathrm{3}}}\:=\:\frac{\boldsymbol{\mathrm{c}}^{\mathrm{3}} }{\mathrm{6}\sqrt{\mathrm{3}}}\:\:\boldsymbol{\mathrm{proved}}. \\ $$

Answered by bobhans last updated on 26/Jun/20

volume of the box = V(x)= (c−2x)^2 x V′(x)= (c−2x)^2 −4(c−2x)x = 0 (c−2x){c−2x−4x} = 0 , x = (1/6)c V_(max) = (c−(1/3)c)^2 .((1/6)c) = (((4c^2 )/9))((c/6)) = ((2c^3 )/(27)) ■

$$\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{box}\:=\:\mathrm{V}\left(\mathrm{x}\right)=\:\left(\mathrm{c}−\mathrm{2x}\right)^{\mathrm{2}} \mathrm{x} \\ $$$$\mathrm{V}'\left(\mathrm{x}\right)=\:\left(\mathrm{c}−\mathrm{2x}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{c}−\mathrm{2x}\right)\mathrm{x}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{c}−\mathrm{2x}\right)\left\{\mathrm{c}−\mathrm{2x}−\mathrm{4x}\right\}\:=\:\mathrm{0}\:,\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\mathrm{c} \\ $$$$\mathrm{V}_{\mathrm{max}} \:=\:\left(\mathrm{c}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{c}\right)^{\mathrm{2}} .\left(\frac{\mathrm{1}}{\mathrm{6}}\mathrm{c}\right)\:=\:\left(\frac{\mathrm{4c}^{\mathrm{2}} }{\mathrm{9}}\right)\left(\frac{\mathrm{c}}{\mathrm{6}}\right) \\ $$$$=\:\frac{\mathrm{2c}^{\mathrm{3}} }{\mathrm{27}}\:\blacksquare \\ $$$$ \\ $$

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