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Show that the surfacearea of a closed cuboid with square base and given volume is minimum, when itis a cube. |
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Answer» Let V be the fixed volume of a closed cuboid with length a, breadth a and height h. Let S be its surface area. Then, `V = (a xx a xx h) or h = (V)/(a^(2))`..(i) Now, `S = 2 (a^(2) + ah + ah) = 2 (a^(2) + 2ah) = 2 (a^(2) + (2V)/(a))`[using (i)] i.e., `S = 2 (a^(2) + (2V)/(a))." ":. (dS)/(da) = 2 (2a - (2V)/(a^(2))) and (d^(2)S)/(da^(2)) = (4 + (8V)/(a^(3)))` Now, `(dS)/(da) = 0 rArr V = a^(3) rArr a xx a xx h = a^(3) rArr h = a` Now, when `h = a`, we have `V = a^(3)` `:. [(d^(2)S)/(da^(2))]_(h = a) = (4 + (8a^(3))/(a^(3))) = 12 gt 0` So, S is minimum when length = a, breadth = a and height = a, i.e., when it is a cube |
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