Posted by **Elizabeth** on Saturday, October 13, 2012 at 10:41pm.

A metal cylindrical container with an open top is to hold 1 cubic foot. If there is no waste in construction, find the dimensions which will require the least amount of material.

- Calculus -
**Reiny**, Sunday, October 14, 2012 at 12:42am
V = πr^2 h

h = 1/(πr^2)

Surface Area (SA)

= bottom + collar of cylinder

= πr^2 + 2πrh

=πr^2 + 2πr(1/(πr^2)

= πr^2 + 2/r

dSA/dr = 2πr - 2/r^2 = 0 for a min of SA

2πr = 2/r^2

r^3 = 1/π

r = 1/π^(1/3) = appr .693 ft

then h = 1/(π(.693^2)) = .683

- Calculus -
**zeeshan amir khan**, Sunday, January 12, 2014 at 8:01am
V = π r ² h = 1ft³

h = 1/π r ²

SA = π r ² + 2 π r h

= π r ² + 2 π r * 1/π r ²

= π r ² + 2 / r

SA ' = 2π r - 2 / r² = 0

2π r = 2 / r²

r³ = 1/π

r = 1/ ³√π ft

h = 1/(π r ²)

= 1/(π (1/ ³√π) ²)

= ³√π ² / π

= 1/ ³√π ft

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