Question

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?

Answer 1

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

Learn more here:

https://brainly.com/question/14316282

https://brainly.com/question/17189053

Answer 2

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:

https://brainly.com/question/16140623?referrer=searchResults