Thursday

April 24, 2014

April 24, 2014

Posted by **Nick** on Friday, April 6, 2012 at 4:24am.

Find the dimensions of the container having the least cost.

- Calc 1 (Optomization) -
**Steve**, Friday, April 6, 2012 at 5:31amarea of base = 2w^2

height = k/2w^2

area of sides = 4*k/2w^2 = 2k/w^2

cost = 6(2w^2) + 10(2k/w^2)

= 12w^2 + 20k/w^2

dc/dw = 24w - 40k/w

dc/dw = 0 when

24w = 40k/w

24w^2 = 40k

w^2 = 5/3 k

w = √(5k/3)

least cost box is thus √(5k/3) x √(5k/3) x 3/10

- Calc 1 - correction -
**Steve**, Friday, April 6, 2012 at 5:35amerror starts here:

dc/dw = 24w - 40k/w

It should be:

dc/dw = 24w - 40k/w^3

24w = 40k/w^3

24w^4 = 40k

w = ∜(5k/3)

least cost box is thus ∜(5k/3) x ∜(5k/3) x √(3k)/(2√5)

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