You want to produce a cylindrical water container with a capacity of 350mL. What dimensions will minimize the amount of material required for the container? Round your answer to the nearest thousandth. Help!?

Question

928 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-09T23:48:25

Hence we have a minimal surface area (amount of material when):

# "radius" = 0.0381911 ... (m) #
# "height" = 0.076382 ... (m)#

It isn't clear what units we should use, so I will omit rounding.

Let us set up the following variables:

# {(r, "Radius", (m)), (y, "Height of container", (m)), (A, "Surface Area of the container", (m^2)), (V, "Volume of the container", (m^3)) :} #

We want to vary the radius #r# such that we minimise #A#, ie find a critical point of #(dA)/(dr)# that is a minimum, so we to find a function #A(r)#

Then the volume is fixed at #350 \ mL = 0.00035 \ m^3#:

Volume of the container is:

# V = pir^2h = 0.00035 #
# => h = 0.00035/(pir^2) #

And, the Surface Area is given by:

# "Side" = 2pirh #
# "Top+Bottom"=2pir^2#

So, the surface area is:

# A= 2pirh + 2pir^2 #

And we can eliminate #h# so that we have #A# as a function of #r# alone (equally we could eliminate #r#).

# :. A = 2pir(0.00035/(pir^2)) + 2pir^2 #
# " " = 0.0007/(r) + 2pir^2 #

Differentiating wrt #r# gives us;

# (dA)/(dr) = - 0.0007/r^2+4pir#

At a critical point, #(dA)/(dr)=0#

# (dA)/(dr) = 0 => - 0.0007/r^2+4pir = 0 #

# :. 4pir = 0.0007/r^2 #
# :. r^3 = 0.0007/(4pi) #
# :. \ \ r = 0.0381911 ... (m) #

With #r=0.0381911# we have:

# A = 0.274932 ... (m^2)#
# h = 0.076382 ... (m)#
# V=0.00035 \ (m^3)#

We should check that this value leads to a minimum (rather than a maximum).

Differentiating the first derivative a second time:

# (d^2A)/(dr^2) = 0.0014/r^3+4pi#
# > 0 " when " r= 0.0381911 ...#

Hence we have a minimal surface area (amount of material when):

# "radius" = 0.0381911 ... (m) #
# "height" = 0.076382 ... (m)#