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
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×l×h=w2h=534
h=534w2

Cost = (Cost of sides) + (Cost of top and bottom)
C=(4×(4w×h))+(8×(2w2))

=(4×(4w×534w2)+(8×(2w2))

=8544w1+16w2

dCdw=0 for critical points
85434w2+32w=0
Assuming w0 we can multiply by w2
and with some simple numeric division:
267+w3=0
and
w=(267)13
=6.44 (approx.)

l=6.44
and h=12.88 (approx.)