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

Hopefully you can visually confirm that a minimum does indeed occur when