Physics, 29.11.2019 07:31 izzyisawesome5232

The volume of a cylindrical tin can with a top and a bottom is to be 16π cubic inches. if a minimum amount of tin is to be used to construct the can, what must be the height, in inches, of the can?

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The required height of the tin can is given by the height at the minimum value of the surface area function of the tin can

The height of the tin can must be 4 inches

Reason:

The given information on the tin can are;

Volume of the tin can = 16·π in.³

The amount of tin to be used = Minimum amount

The height of the can in inches required

Solution;

Let h, represent the height of the can, we have;

The surface area of the can, S.A. = 2·π·r² + 2·π·r·h

The volume of the can, V = π·r²·h

Where;

r = The radius of the tin can

h = The height of the tin can

Which gives;

16·π = π·r²·h

16 = r²·h

$h = \dfrac{16}{r^2}$

Which gives;

$S.A. = 2 \cdot \pi \cdot r^2 + 2 \cdot \pi \cdot r \cdot \dfrac{16}{r^2} = 2 \cdot \pi \cdot r^2 + \pi \cdot \dfrac{32}{r}$

When the minimum amount of tin is used, we have;

$\dfrac{d(S.A.)}{dx} =0 = \dfrac{d}{dx} \left( 2 \cdot \pi \cdot r^2 + \pi \cdot \dfrac{32}{r} \right) = \dfrac{4 \cdot \pi \cdot \left (r^3-8 \right)}{r^2}$

Therefore;

$\dfrac{4 \cdot \pi \cdot \left (r^3-8 \right)}{r^2} = 0$

4·π·(r³ - 8) = r² × 0

r³ = 8

r = 2

The radius of the tin can, r = 2 inches

The

$h = \dfrac{16}{2^2} = 4$

The height of the tin can, h = 4 inches

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Answer from: Asrys

The required height of can is 4 inches.

Given data:

The volume of cylindrical tin with top and bottom is, $V = 16 \pi \;\rm in^{3}$ .

The volume of tin is,

$V = \pi r^{2}h\\16 \pi = \pi r^{2}h\\$

here, r is the radius of can and h is the height of can.

$16 =r^{2}h\\\\h=\dfrac{16}{r^{2}}$

The surface area of tin is,

$SA = 2\pi r(r+h)\\SA = 2\pi r(r+\dfrac{16}{r^{2}})\\SA = 2\pi r^{2}+\dfrac{32 \pi}{r})$

For minimum amount of tin used, we have,

$\dfrac{d(SA)}{dr} =0\\\dfrac{ d(2\pi r^{2}+\dfrac{32 \pi}{r})}{dr} = 0\\4\pi r-\dfrac{32 \pi}{r^{2}}=0\\4\pi r=\dfrac{32 \pi}{r^{2}}\\r^{3}=8\\r =2$

So, height of can is,

$h=\dfrac{16}{2^{2}}\\h=4$

Thus, the required height of can is 4 inches.

Learn more about the curved surface area here:

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Answer from: Quest

Hello how was your day? ? i hope it was good

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

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The volume of a cylindrical tin can with a top and a bottom is to be 16π cubic inches. if a minimum...

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