optimization A large bin for holding heavy material must be in the shape of a box with an open top and a square base. The base will cost 6 dollars a square foot and the sides will cost 8 dollars a...
optimization
A large bin for holding heavy material must be in the shape of a box with an open top and a square base. The base will cost 6 dollars a square foot and the sides will cost 8 dollars a foot. If the volume must be 160 cubic feet. Find the dimensions that will minimize the cost of the box's construction. Find base and each side
(Leave your answers to 3 decimal places.)
Base :
Each side :
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You should use the notations such that: x for the lengths of square base and y for the height of box, hence, evaluating the volume of the box yields:
`V = x^2*y => 160 = x^2*y => y = 160/x^2`
You need to evaluate the total surface area of box such that:
`A(x) = x^2 + 4x*y`
You should substitute `160/x^2` for y in equation of `A(x)` such that:
`A(x) = x^2 + 4x*160/x^2 => A(x) = x^2 + 640/x`
`A(x) = (x^3+160)/x`
You need to optimize the dimensions of the box, hence, you need to find the solutions to the equation `A'(x)=0` such that:
`A'(x) = ((x^3+160)/x)' =>A'(x) = (3x^2*x - x^3 - 160)/x^2`
`A'(x) = (2x^3 - 160)/x^2`
`A'(x)=0 <=> 2x^3 - 160 = 0 => x^3 - 80 = 0`
Using the formula `(a-b)(a^2+ab+b^2)=a^3-b^3` yields:
`x^3 - 80 = (x - root(3)80)(x^2 + xroot(3)80 + root(3)(80^2))`
`x^3 - 80 = 0 => x - root(3)80 = 0 => x = root(3)80`
`x^2 + xroot(3)80 + root(3)(80^2) != 0 for x in R`
Hence the base `x = root(3)80 ` and the height `y = 160/(root(3)(80^2))`
Hence, evaluating the dimensions that minimize the cost of the box yields `x = root(3)80 feet ` and `y = 160/(root(3)(80^2)) feet.`
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