# A cylindrical can, open at the top, is to hold 510 cm3 of liquid. Find the height and radius that minimize the amount of material needed to ?

A cylindrical can, open at the top, is to hold 510 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.

### 3 Answers

- RichardLv 74 weeks ago
Let the radius be 'r' and its height be 'h'

The area of the base is πr²

The area of the wall of the can is 2πrh

The volume is πr²h = 510 cm³

So h = 510 / πr²

The area of can's material = πr² + 2πrh

Substitute for h

= πr² + (2πr * 510 / πr²)

= πr² + 1020/r

differentiate with respect to r

d(πr² + 1020/r) / dr = 2πr - 1020/r²

equate this to 0 for the maximum or minimum.

2πr = 1020/r²

r³ = 510 / π = 162.338

r = 5.45515 cm

h = 510 / πr² = 510 / 93.48962 = 5.45515 cm

r = h = 5.45515 cm.

- Geeganage WLv 54 weeks ago
Let the height be "h".

The radius, r, gives; 510 = π(r^2)h, h = 510/[π(r^2)],

Area, say, A.

A = π(r^2) +2πrh = π(r^2) +1020/r.

dA/dr = 2πr -1020/r^2,

dA/dr = 0 when r = (510/π)^(1/3).

d^2A/dx^2 = 2π +3060/(r^3) >0 for all r.

When r = (510/π)^(1/3), A becomes minimum.

h = 510/[π(r^2)] = (510/𝜋)^(1/3),

Therefore, to satisfy the condition,

r = (510/𝜋)^(1/3),

h = (510/𝜋)^(1/3), r = h = (510/pi)^(1/3).

- 4 weeks ago
Given a specific volume, a can with an open top will use the smallest amount of material when the radius and the height are the same. Let's show how:

V = pi * r^2 * h

V / (pi * r^2) = h

A = pi * r^2 + 2 * pi * r * h

A = pi * r^2 + 2 * pi * r * V / (pi * r^2)

A = pi * r^2 + 2V/r

dA/dr = 2pi * r - 2V/r^2

dA/dr = 0

0 = 2pi * r - 2V/r^2

0 = pi * r - V/r^2

V/r^2 = pi * r

V = pi * r^3

V/pi = r^3

(V/pi)^(1/3) = r

V / (pi * r^2) = h

h = V / (pi * (V/pi)^(2/3))

h = V * pi^(2/3) / (pi * V^(2/3))

h = V^(1/3) / pi^(1/3)

h = (V/pi)^(1/3)

h = r

Okay, with that knowledge, we can find h and r for any can like this one.

r = h = (510/pi)^(1/3)

thank you!!!! :)