Question #3498f
1 Answer
The smallest cost occurs when we have a radius of
Explanation:
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
Then the volume is fixed at
# 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
# :. C=16pir(20/r^2) + 20pir^2 #
# :. C=320pi/r + 20pir^2 #
Differentiating wrt
# :. (dC)/(dr)=(320pi)(-1/r^2) + 40pir #
# :. (dC)/(dr)=(-320pi)/r^2 + 40pir #
At a critical point,
# :. (-320pi)/r^2 + 40pir = 0 #
# :. -320+40r^3=0 #
# :. r^3=320/40 #
# :. r^3=8 #
# :. r = 2 #
With
# 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
graph{(320pi)/x + (20pi)*x^2 [-5, 5, -100, 1000]}
Hopefully you can visually confirm that a minimum does indeed occur when