How do you find the dimensions of the box that minimize the total cost of materials used if a rectangular milk carton box of width w, length l, and height h holds 534 cubic cm of milk and the sides of the box cost 4 cents per square cm and the top and bottom cost 8 cents per square cm?

1 Answer
Alan P.
Mar 11, 2015

Note that varying the length and width to be other than equal reduces the volume for the same total (length + width); or, stated another way, #w = l# for any optimal configuration.

Using given information about the Volume, express the height (#h#) as a function of the width (#w#).

Write an expression for the Cost in terms of only the width (#w#).

Take the derivative of the Cost with respect to width and set it to zero to determine critical point(s).

Details:
Volume #= w xx l xx h = w^2 h = 534#
#rarr h = (534)/(w^2)#

Cost = (Cost of sides) + (Cost of top and bottom)
#C= (4 xx (4w xx h)) + ( 8 xx (2 w^2))#

#= (4 xx (4w xx (534)/(w^2)) + (8 xx (2 w^2))#

#= 8544 w^(-1) + 16 w^2#

#(d C)/(dw) = 0# for critical points
# - 85434 w^(-2) + 32 w = 0#
Assuming #w != 0# we can multiply by #w^2#
and with some simple numeric division:
#- 267 + w^3 = 0#
and
#w = (267)^(1/3)#
#= 6.44# (approx.)

#rarr l = 6.44#
and #h = 12.88# (approx.)

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