Optimization- Word Problem, minimum amount of material to constuct

**franklen** · November 10th 2007, 08:56 PM

A grain silo has the shape of a right circular cylinder surmounted by a hemisphere. If the silo is to have a capacity of

, find the radius and height of the silo that requires the least amount of material to construct.

Hint: The volume of the silo is , and the surface area (including the floor) is

... I need to find "r" and "h"

**Soroban** · November 11th 2007, 03:47 AM

Hello, franklen!

A grain silo has the shape of a right circular cylinder surmounted by a hemisphere.
If the silo is to have a capacity of $614\pi$ ft³, find the radius $r$ and height $h$ of the silo
that requires the least amount of material to construct.

Hint: The volume of the silo is: . $V \:=\;\pi r^2h + \frac{2}{3}\pi r^3$ .[1]
. . . and the surface area (including the floor) is: . $S \:=\:3\pi r^2 + 2\pi rh$ .[2]

They gave us the necessary formulas . . .

From [1], we have: . $V \:=\:\pi r^2h + \frac{2}{3}\pi r^3 \:=\:614\pi \quad\Rightarrow\quad h \:=\:614r^{-2} + \frac{2}{3}r$ .[3]

Substitute into [2]: . $S \;=\;3\pi r^2 + 2\pi r\left(614r^{-2} + \frac{2}{3}r\right)$
. . which simplifies to: . $S \;=\;\frac{13}{3}\pi r^2 + 1228\pi r^{-1}$

Differentiate and equate to zero: . $S' \;=\;\frac{26}{3}\pi r - 1228\pi r^{-2} \;=\;0$

Muliply by $3r^2\!:\;\;26\pi r^3 - 3684\pi \;=\;0$

Then we have: . $r^3 \;=\;\frac{3684\pi}{26\pi} \;=\;\frac{1842}{13} \quad\Rightarrow\quad \boxed{r \;=\;\sqrt[3]{\frac{1842}{13}}}$

From .[3], we have: . $h \;=\;\frac{1842 + 2r^3}{3r^2}$

Substitute: . $h \;=\;\frac{1842 + 2\left(\frac{1842}{13}\right)}{3\left(\frac{1842}{ 13}\right)^{\frac{2}{3}}}$

. . and after a LOT of algebra, we get: . $\boxed{h \;=\;5\sqrt[3]{\frac{1842}{13}}}$

Fascinating . . . the height is exactly five times the radius.
.

**franklen** · November 11th 2007, 09:15 AM

hmm... it says that those answers are incorrect?

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