Question

cylindrical container is to be constructed to be open at the top with a volume of 27π cubic meters using the least amount of material. Find the radius and height of the cylinder which will require the least amount of material to construct.

Answer 1

radius = 3 m

Height = 3 m

Step-by-step 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

Answer 2

radius comes out to be 3 m

height of the cylinder comes out to be 3m

Step-by-step 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 (27)/(r) + \pi r^2`

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

`\frac{\mathrm{d} s}{\mathrm{d} r}=54\pi (-1)/(r^2)+2\pi r`

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

for least amount of material requirement.

`(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