What dimensions of the rectangle will result in a cylinder of maximum volume if you consider a rectangle of perimeter 12 inches in which it forms a cylinder by revolving this rectangle about one of its edges?

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What dimensions of the rectangle will result in a cylinder of maximum volume if you consider a rectangle of perimeter 12 inches in which it forms a cylinder by revolving this rectangle about one of its edges?

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Answer 1 · 2015-02-05T22:17:41

A cylinder's volume, $V = π r^{2} h$ $V=pir^2h$
The width of the rectangle, which forms the circumference, $w = 2 π r$ $w=2pir$
graph{x+y=12 [-2.49, 25.99, -1.51, 12.74]}
The graph shows possible lengths for the sides of the rectangle if they must add up to 12 inches ( $w$ $w$ is horizontal and $h$ $h$ is the vertical axis).
At what point along that curve is $\frac{w^{2}}{4 π^{2}} \cdot h = r^{2} h$ $w^2/(4pi^2)*h=r^2h$ the greatest?

graph{(12x-x^2)/(4pi) [-0.89, 13.35, -1.027, 6.1]}

Since $r = \frac{w}{2 π} = \frac{12 - h}{2 π}$ $r=w/(2pi)=(12-h)/(2pi)$ , you can plot a graph of the volume for different values of $h$ $h$ with the equation $y = π \cdot \frac{12 - x}{4 π^{2}} \cdot x = \frac{12 x - x^{2}}{4 π}$ $y=pi*(12-x)/(4pi^2)*x=(12x-x^2)/(4pi)$ . Differentiating to find the maximum:

$\frac{d y}{d x} = \frac{12}{4 π} - \frac{2 x}{4 π} = \frac{3}{π} - \frac{x}{2 π} = 0$ $dy/dx=12/(4pi)-(2x)/(4pi)=3/pi-x/(2pi)=0$

$x = h = 6$ $x=h=6$

Since $h = 6$ $h=6$ , $w$ $w$ must also equal $6$ $6$

$V = π r^{2} \cdot h = π \cdot {(\frac{6}{2 π})}^{2} \cdot 6 = 533 π = 1674$ $V=pir^2*h=pi*(6/(2pi))^2*6=533pi=1674$

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What dimensions of the rectangle will result in a cylinder of maximum volume if you consider a rectangle of perimeter 12 inches in which it forms a cylinder by revolving this rectangle about one of its edges?

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