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?

Question

Emily Ammerman · Accepted 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).

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 #A# as a function of one variable (#l# or #w#) and hopefully be able to minimise #A# wrt that variable. 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 #(d^2A)/(dw^2) > 0#. # \ \ \ \ \ \ (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).

Charles Anctil · Answer

undefined To minimize the total surface area of the box, the dimensions of the rectangular base should be as close as possible, while still satisfying the volume constraint. Let the width of the base be $ w $ feet. Then the length of the base is $ 5w $ feet, and the height of the box is $ h $ feet.

Given that the volume $ V $ of the box is 10 cubic feet, we have:

$$ V = \text{length} \times \text{width} \times \text{height} $$
$$ 10 = (5w) \times w \times h $$
$$ 10 = 5w^2h $$
$$ h = \frac{10}{5w^2} $$
$$ h = \frac{2}{w^2} $$

The total surface area $ A $ of the box is given by:

$$ A = 2lw + 2lh + lw $$
$$ A = 2(5w)w + 2(5w)\left(\frac{2}{w^2}\right) + (5w)w $$
$$ A = 10w^2 + \frac{20}{w} + 5w^2 $$
$$ A = 15w^2 + \frac{20}{w} $$

To find the minimum value of $ A $, we differentiate $ A $ with respect to $ w $ and set the derivative equal to zero:

$$ \frac{dA}{dw} = 30w - \frac{20}{w^2} $$
$$ 0 = 30w - \frac{20}{w^2} $$
$$ 20 = 30w^3 $$
$$ w^3 = \frac{2}{3} $$
$$ w = \left(\frac{2}{3}\right)^{\frac{1}{3}} $$

Substituting the value of $ w $ back into the expressions for $ l $ and $ h $, we find:

$$ l = 5w = 5\left(\frac{2}{3}\right)^{\frac{1}{3}} $$
$$ h = \frac{2}{w^2} = \frac{2}{\left(\left(\frac{2}{3}\right)^{\frac{1}{3}}\right)^2} $$

So, the dimensions of the box that minimize the total surface area are approximately:

$$ \text{Width: } \left(\frac{2}{3}\right)^{\frac{1}{3}} \text{ feet} $$
$$ \text{Length: } 5\left(\frac{2}{3}\right)^{\frac{1}{3}} \text{ feet} $$
$$ \text{Height: } \frac{2}{\left(\left(\frac{2}{3}\right)^{\frac{1}{3}}\right)^2} \text{ feet} $$