# A canned fruit producer wishes to minimise the area of sheet metal used in manufacturing cans of given volume. Find the ratio to height of the desired can. (The can has closed ends)

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Let shape of the cane be cylindrical . Also assume height and radius of the cane be h and r respctively.

Thus volume of the cane

`V=pir^2h` (i)

Thus

`h=V/(pir^2)` (ii)

total surface area ( sheet required )

`S=2pir(r+h)` (iii)

substitute h frm (ii) in (iii)

`S=2pir^2+2pir(V/(pir^2))`

`S=2pir^2+(2V)/r`

`(dS)/(dr)=4pir-2V/r^2`

`for minima`

`(dS)/(dr)=0=4pir-(2V)/r^2`

`r=(V/(2pi))^(1/3)`

`(d^2S)/(dr^2)=4pi+(4V)/(r^3)`

`(d^2S)/(dr^2)}_{r=(V/(2pi))^(1/3)}>0`

`` Thus

`r=(V/(2pi))^(1/3)` will provide minimum surface area of the cane.