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A-rectangular-box-open-at-the-top-is-to-have-a-volume-of-32-cube-feet-What-must-be-the-dimensions-so-that-the-total-surface-is-a-minimum-

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Question Number 144532 by EDWIN88 last updated on 26/Jun/21
A rectangular box,open at the top is to have a volume of 32 cube feet What must be the dimensions so that the total surface is a minimum?
$$\mathrm{A}\:\mathrm{rectangular}\:\mathrm{box},\mathrm{open}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{top}\:\mathrm{is}\:\mathrm{to}\:\mathrm{have}\:\mathrm{a}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{32}\:\mathrm{cube}\:\mathrm{feet} \\ $$$$\mathrm{What}\:\mathrm{must}\:\mathrm{be}\:\mathrm{the}\:\mathrm{dimensions} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{total}\:\mathrm{surface}\:\mathrm{is}\:\mathrm{a}\:\mathrm{minimum}? \\ $$
Answered by liberty last updated on 26/Jun/21
volume of box = V=xyz = 32 and z = ((32)/(xy)) surface area of box =S=xy+2yz+2xz { (((∂S/∂x)=y−((64)/x^2 )=0 when x^2 y=64)),(((∂S/∂y)=x−((64)/y^2 )=0 when xy^2 =64)) :} so we get y=x and x^3 =64 or x=y=4 and z=2 for x=y=4 ,Δ=S_(xx) S_(yy) −S_(xy) ^2 Δ = (((128)/x^3 ))(((128)/y^3 ))−1>0 and S_(xx) = ((128)/x^3 )>0 hence it follows that the dimensions 4 feet × 4 feet ×2 feet give the minimum surface
$$\mathrm{volume}\:\mathrm{of}\:\mathrm{box}\:=\:\mathrm{V}=\mathrm{xyz}\:=\:\mathrm{32} \\ $$$$\:\mathrm{and}\:\mathrm{z}\:=\:\frac{\mathrm{32}}{\mathrm{xy}} \\ $$$$\mathrm{surface}\:\mathrm{area}\:\mathrm{of}\:\mathrm{box}\:=\mathrm{S}=\mathrm{xy}+\mathrm{2yz}+\mathrm{2xz} \\ $$$$\:\begin{cases}{\frac{\partial\mathrm{S}}{\partial\mathrm{x}}=\mathrm{y}−\frac{\mathrm{64}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{0}\:\mathrm{when}\:\mathrm{x}^{\mathrm{2}} \mathrm{y}=\mathrm{64}}\\{\frac{\partial\mathrm{S}}{\partial\mathrm{y}}=\mathrm{x}−\frac{\mathrm{64}}{\mathrm{y}^{\mathrm{2}} }=\mathrm{0}\:\mathrm{when}\:\mathrm{xy}^{\mathrm{2}} =\mathrm{64}}\end{cases} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{get}\:\mathrm{y}=\mathrm{x}\:\mathrm{and}\:\mathrm{x}^{\mathrm{3}} =\mathrm{64}\:\mathrm{or}\: \\ $$$$\mathrm{x}=\mathrm{y}=\mathrm{4}\:\mathrm{and}\:\mathrm{z}=\mathrm{2} \\ $$$$\mathrm{for}\:\mathrm{x}=\mathrm{y}=\mathrm{4}\:,\Delta=\mathrm{S}_{\mathrm{xx}} \mathrm{S}_{\mathrm{yy}} −\mathrm{S}_{\mathrm{xy}} ^{\mathrm{2}} \\ $$$$\Delta\:=\:\left(\frac{\mathrm{128}}{\mathrm{x}^{\mathrm{3}} }\right)\left(\frac{\mathrm{128}}{\mathrm{y}^{\mathrm{3}} }\right)−\mathrm{1}>\mathrm{0}\: \\ $$$$\mathrm{and}\:\mathrm{S}_{\mathrm{xx}} =\:\frac{\mathrm{128}}{\mathrm{x}^{\mathrm{3}} }>\mathrm{0}\:\mathrm{hence}\:\mathrm{it}\:\mathrm{follows} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{dimensions}\:\mathrm{4}\:\mathrm{feet}\:×\:\mathrm{4}\:\mathrm{feet}\:×\mathrm{2}\:\mathrm{feet} \\ $$$$\mathrm{give}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{surface} \\ $$

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