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).