Respuesta :
The dimensions of the cylinder that has the maximum volume can be
found be finding the maximum value of the volume function.
- [tex]\mathrm{The \ \underline{height \ of \ the \ cylinder \ of \ maximum \ volume \ is \ \dfrac{8}{\sqrt{3} }}}[/tex]
- [tex]\mathrm{The \ \underline{radius \ of \ the \ cylinder \ of \ maximum \ volume \ is \ \dfrac{8 \cdot \sqrt{6} }{3}}}[/tex]
Reasons:
The known parameter of the hemisphere are;
The radius of the hemisphere = 8
Required;
The radius and height of the maximum volume of the inscribed cylinder.
Solution:
Let h represent the height of the cylinder, and let r represent the radius of
the cylinder, we have;
r² + h² = 8²
Therefore;
r² = 8² - h² = 64 - h²
r² = 64 - h²
The volume of the cylinder, V = π·r²·h = π·(64 - h²)·h
V(h) = π·(64·h - h³)
At the maximum volume, we have;
[tex]\dfrac{dV(h)}{dh} = \dfrac{d}{dh} \left(\pi \cdot \left(64 \cdot h - h^3 \right) \right) = \pi \cdot \left(64 - 3 \cdot h^2 \right) = 0[/tex]
64 - 3·h² = 0
64 = 3·h²
[tex]h =\sqrt{\dfrac{64}{3} } = \dfrac{8}{\sqrt{3} }[/tex]
[tex]\underline{\mathrm{The \ height \ of \ the \ cylinder, \ h} = \dfrac{8}{\sqrt{3} }}[/tex]
[tex]r^2 = 64 - \dfrac{64}{3} = \dfrac{128}{3}[/tex]
[tex]r =\sqrt{\dfrac{128}{3}} = \dfrac{8 \cdot \sqrt{6} }{3}[/tex]
[tex]\underline{\mathrm{The \ radius \ of \ the \ cylinder, \ } r = \dfrac{8 \cdot \sqrt{6} }{3}}[/tex]
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