# An open top box is to have a rectangular base for which the length is 5 times the width and a volume of 10 cubic feet. It's five sides are to have as small a total surface area as possible. What are the sides?

##### 1 Answer

The box should have the following dimensions (2dp):

# {(1.34, "Width of box (feet)"), (6.69, "Length of box (feet)"), (1.13, "Height of box (feet)") :} #

I'll leave as an exercise conversion to inches (if required).

#### Explanation:

Let us set up the following variables:

# {(w, "Width of box (feet)"), (l, "Length of box (feet)"), (h, "Height of box (feet)"), (A, "Total Surface Area (sq feet)"), (V, "Total Volume (cubic feet)") :} #

Our aim is to get the total surface area

The width/length constraint gives us:

# l=5w #

The volume constraint gives us:

# \ \ \ \ \ \ V=wlh #

# :. 10 =w(5w)h #

# :. 10 = 5w^2h #

# :. \ h = 2/w^2 #

And the surface area of the box is given by:

# A = ("area base") + 2 xx ("area ends") + 2 xx ("area sides") #

# \ \ \ = wl + 2(wh) + 2(lh) #

# \ \ \ = w(5w) + 2(w)(2/w^2) + 2(5w)(2/w^2) #

# \ \ \ = 5w^2 + 4/w + 20/w #

# \ \ \ = 5w^2 + 24/w #

Which we can differentiate to find the critical points

# (dA)/(dw) = 10w-24/w^2 #

# (dA)/(dw) = 0 => 10w-24/w^2 = 0#

# :. 10w=24/w^2#

# :. w^3=2.4 #

# :. w=root(3)(2.4) = 1.338865 ...#

With this dimension we have:

# l = 5w = 6.694329 ... #

# h = 2/w^2 = 1.115721 ... #

# A = 26.888428 ... #

# V = 10 #

We should check that this value leads to a minimum (rather than a maximum) surface area by checking that

# \ \ \ \ \ \ (dA)/(dw) = 10w-24/w^2 #

# :. (d^2A)/(dw^2) = 10+8/w^3 > 0 " when "w=1.338865 ...#

Hence, The box should have the following dimensions (2dp):

# {(1.34, "Width of box (feet)"), (6.69, "Length of box (feet)"), (1.13, "Height of box (feet)") :} #

I'll leave as an exercise conversion to inches (if required).