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-

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 ?

Answered by liberty last updated on 26/Jun/21

volume of box = V = xyz = 32

and z = \frac{32}{xy}

surface area of box = S = xy + 2 yz + 2 xz

{\begin{cases} \frac{\partial S}{\partial x} = y - \frac{64}{x^{2}} = 0 when x^{2} y = 64 \\ \frac{\partial S}{\partial y} = x - \frac{64}{y^{2}} = 0 when {xy}^{2} = 64 \end{cases}

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 = (\frac{128}{x^{3}}) (\frac{128}{y^{3}}) - 1 > 0

and S_{xx} = \frac{128}{x^{3}} > 0 hence it follows

that the dimensions 4 feet \times 4 feet \times 2 feet

give the minimum surface

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