An open box with a square bottom with a volume of 96 cubic inches is to be constructed. The bottom material costs three times as much per square inch as the side material. What dimensions will minimize the cost of the box?

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An open box with a square bottom with a volume of 96 cubic inches is to be constructed. The bottom material costs three times as much per square inch as the side material. What dimensions will minimize the cost of the box?

Mathematics
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volume of a box = length*width*height; and since its a square bottom, we could get away with: V = w^2 h = 96 the area of the sides of the box amount to 4 sides of w*h; and a bottom of w^2 4wh + w^2
lets find a good enough value for "h" to plug into this: We do that by using the volume 96 = w^2 h, h = 96/w^2
4w(96/w^2) + w^2(96/w^2) = A 384/w + 96 = A this is how much we have altogether in materials

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the amount of material in the bottom cost 3times as much as the sides, so amount times price = cost..
an open box with the square bottom of having length 2 inches and the whole box having height of 24 inches
that is one possible scenario, yes :) but many dimensions can have the same area... i think
i use differentiation to get the minimum cost
and my equation has an error in it, we dont multiply height to the second term 4w(96/w^2) + w^2 = A 384/w + w^2 = A yes, that is how you would optimize the problem :)
the bottom costs 3times as much....so lets assume that the price per side area is 1 and the price for the bottom would be 3 384/w + 3x^2 = total cost
derive and we get: -384/w^2 + 6w = C' -384 + 6w^3 ------------ = 0 w^2 6w^3 = 384 w^3 = 64 w = cbrt(64) w = 4 but i need to recheck my work ;)
it checks out in my books... the base is 4 by 4 16(h) = 96 h = 96/16 h = 6 4x4x6 would be the best option. lets compare that to 2x2x24 (2)(2)(3) + 4(2)(24) 12 + 192 = $204 -------------------- (4)(4)(3) + 4(4)(6) 48 + 96 = $144
I would go with the 4x4x6 measurements :)
haha yep u're right. I made a glaring mistake, argh
I was sweating it :) til I got to the end lol

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