A cylindrical can, open at the top, is to hold 200 cm3 of liquid. Find the height and radius that minimize the amount of material needed to?

Question

A cylindrical can, open at the top, is to hold 200 cm3 of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can. Enter your answer with rational exponents, and use pi to represent 𝜋

Radius:

Height:

Answer 1

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The volume of the cylinder is:

v = (π.d²/4) * h → where d is the diameter of the cylinder and where h is the height of the cylinder

v = (π.d²/4) * h → given that the volume is 200 cm³

(π.d²/4) * h = 200

π.d² * h = 800

h = 800/(π.d²)

The volume of the required material to build the cylinder is proportional to the surface area of the cylinder, if the thickness is constant. The surface area (including the bottom) of the cylinder is:

s = [curved surface area] + [bottom surface area]

s = [π.d * h] + [π.d²/4] → we've just seen that: h = 800/(π.d²)

s = [π.d * 800/(π.d²)] + [π.d²/4]

s = [800/d] + [π.d²/4]

s = (3200 + π.d³)/(4d) ← this is a function of d

You can obtain the minimum of a function when its derivative is zero.

The function s looks like (u/v), so its derivative looks like: [(u'.v) - (v'.u)]/v² → where:

u = 3200 + π.d³ → u' = 3π.d²

v = 4d → v' = 4

s' = [(u'.v) - (v'.u)]/v²

s' = [(3π.d² * 4d) - 4.(3200 + π.d³)] / (4d)²

s' = [12π.d³ - 12800 - 4π.d³] / (16d²)

s' = [8π.d³ - 12800] / (16d²)

s' = [π.d³ - 1600] / (2d²) → then you solve for d, the equation: s' = 0

[π.d³ - 1600] / (2d²) = 0 → where: d ≠ 0

π.d³ - 1600 = 0

π.d³ = 1600

d³ = 1600/π

d = (1600/π)^(1/3)

→ d ≈ 8 cm

→ r ≈ 4 cm

Recall: h = 800/(π.d²)

→ h ≈ 4 cm

…and you can see that you can minimize the material when: r = h

The volume of the required material to build the cylinder is proportional to the surface area:

s = (3200 + π.d³)/(4d)

s = (3200 + π.{1600/π})/[4 * (1600/π)^(1/3)]

s = (3200 + 1600)/[4 * (1600/π)^(1/3)]

s = 4800/[4 * (1600/π)^(1/3)]

s = 1200/(1600/π)^(1/3)

→ s ≈ 150.3 cm²

…then you multiply this value to the thickness of the material to obtain the volume.

Answer 2

V= π r² h= 200

S= π r² + 2π r h

h= 200/(π r²)

S= πr² + 400/r

dS/dr= 2π r - 400/r= 0

2π r= 400/r²; r^3= 200/π; r = (200/π)^(1/3)≈ 3.99 cm

h= 200/(π 3.99²)≈ 3.99 cm