# 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?

Nathanael Perkins 2022-09-27 Answered
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?
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## Answers (1)

baselulaox
Answered 2022-09-28 Author has 8 answers
Note that varying the length and width to be other than equal blackuces 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 $=w×l×h={w}^{2}h=534$
$\to h=\frac{534}{{w}^{2}}$
Cost = (Cost of sides) + (Cost of top and bottom)
$C=\left(4×\left(4w×h\right)\right)+\left(8×\left(2{w}^{2}\right)\right)$
$=\left(4×\left(4w×\frac{534}{{w}^{2}}\right)+\left(8×\left(2{w}^{2}\right)\right)$
$=8544{w}^{-1}+16{w}^{2}$
$\frac{dC}{dw}=0$ for critical points
$-85434{w}^{-2}+32w=0$
Assuming $w\ne 0$ we can multiply by ${w}^{2}$ and with some simple numeric division:
$-267+{w}^{3}=0$
and
$w={\left(267\right)}^{\frac{1}{3}}$
=6.44 (approx.)
$\to l=6.44$
and h=12.88 (approx.)
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