Question #3498f

1 Answer
Dec 2, 2016

The smallest cost occurs when we have a radius of #2#m and a height of #5#m, leading to a cost of $#753.98#.

Explanation:

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Let us set up the following variables:

# {(r, "Radius (m)"), (y, "Height of can (m)"), (C, "Cost of the can ($)") :} #

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

Then the volume is fixed at #20pi# # m^3#:

# pir^2h = 20pi #
# :. r^2h = 20 #
# :. h = 20/r^2 #

And, the Surface Area are given by:

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

So, the total cost is:

# C=2pirh*8 + 2pir^2*10 #
# :. C=16pirh + 20pir^2 #

And we can eliminate #h# so that we have #C# as a function of #r# alone (equally we could eliminate #r# and have #C=f(h)# and find #h# st #(dC)/(dh)=0#).

# :. C=16pir(20/r^2) + 20pir^2 #
# :. C=320pi/r + 20pir^2 #

Differentiating wrt #r# gives us;

# :. (dC)/(dr)=(320pi)(-1/r^2) + 40pir #
# :. (dC)/(dr)=(-320pi)/r^2 + 40pir #

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

# :. (-320pi)/r^2 + 40pir = 0 #
# :. -320+40r^3=0 #
# :. r^3=320/40 #
# :. r^3=8 #
# :. r = 2 #

With #r=2# we have:

# C=(320pi)/2 + 20pi*4 #
# :. C=160pi + 80pi #
# :. C=240pi ~~ 753.98# (2dp)

And,

# h=20/4=5#

We should check that this value leads to a minimum (rather than a maximum) cost. As the size of the can is finite this should really be intuitive. We could calculate the second derivative and verify that #(d^2C)/(dr)^2 > 0# when #r=2# Instead I will just use the graph #C=320pi/r + 20pir^2#

graph{(320pi)/x + (20pi)*x^2 [-5, 5, -100, 1000]}

Hopefully you can visually confirm that a minimum does indeed occur when #r=2#