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Serjik [45]
3 years ago
10

cylindrical container is to be constructed to be open at the top with a volume of 27π cubic meters using the least amount of mat

erial. Find the radius and height of the cylinder which will require the least amount of material to construct.
Physics
2 answers:
svetlana [45]3 years ago
6 0

Answer:

radius = 3 m

Height = 3 m

Explanation:

Let r be the radius of the cylinder and h be the height.

Voluem of cylinder is given by

V = π x r² x h

27 π = π x r² x h

h = 27 / r²      .... (1)

Material rquired to make open top is curved surface area and the area of base

S = π r² + 2 π r h

S = π r² + 2 π r x 27 / r²              (from equation (1)

S = π r² + 54 π / r

Differentiate with respect to r

dS / dr = 2 x π x r - 54 π / r²

It should be zero for maxima and minima

2 x π x r - 54 π / r² = 0

r = 3 m

Put in equation (1), we get

h = 27 / (3 x 3) = 3 m

Differentiate dS / dr again

d²S / dr² = 2 π + 108 π / r³    = Positive

So, the surface area S is minimum for r = 3 m and h = 3 m

Llana [10]3 years ago
3 0

Answer:

radius comes out to be 3 m

height of the cylinder comes out to be 3m

Explanation:

given

volume of cylinder = 27π m³

π r² h = 27π

   r² h = 27.............(1)

surface area of cylinder open at the top

S = 2πrh + π r²

S = 2\pi \dfrac{27}{r} + \pi r^2

\frac{\mathrm{d} s}{\mathrm{d} r}=\frac{\mathrm{d}}{\mathrm{d} r} (2\pi \dfrac{27}{r} + \pi r^2)

\frac{\mathrm{d} s}{\mathrm{d} r}=54\pi \dfrac{-1}{r^2}+2\pi r

\frac{\mathrm{d} s}{\mathrm{d} r}=0

for least amount of material requirement.

\dfrac{54\pi }{r^2} = 2\pi r\\r=3m

hence radius comes out to be 3 m

for height put the value in the equation 1

so, height of the cylinder comes out to be 3m

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