Asked by May
Find the most economical proportions for a covered box with a volume of 473cm³ whose base is rectangle with one side three times as long as the others
Answers
Answered by
oobleck
so if the base is x by 3x cm, we have
3x^2 h = 473
and thus the height h = 473/(3x^2)
Not sure what you mean by "most economical" but if it means using the least material, then the area a is
a = 2(3x^2 + 2*xh + 2*3xh) = 2(3x^2 + 8x(473/(3x^2)))
a = 2(3x^2 + 3784/(3x))
da/dx = 2(6x - 3784/(3x^2))
so now just find where da/dx = 0 for the minimum value of a
3x^2 h = 473
and thus the height h = 473/(3x^2)
Not sure what you mean by "most economical" but if it means using the least material, then the area a is
a = 2(3x^2 + 2*xh + 2*3xh) = 2(3x^2 + 8x(473/(3x^2)))
a = 2(3x^2 + 3784/(3x))
da/dx = 2(6x - 3784/(3x^2))
so now just find where da/dx = 0 for the minimum value of a
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