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Show that the total surface area of a closed cuboidwith square base and given volume is minimum, when it is a cube. |
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Answer» Volume of a cuboid witha square base `V= l**l**h` Here, `l` is length and breadth and `h` is the height. `:. V = l^2h => h = V/l^2` Now, total surface area of the cuboid `S = 2(l**l+l**h+l**h) ` `=> S = 2l^2+4lh` `=> S = 2l^2+4l(V/l^2)` `=> S = 2l^2+(4V)/l` Now, surface area will be minimum when `(dS)/(dl)` is `0`. `=>(dS)/(dl) = 4l-(4V)/l^2` For minimum surface area, `=> 4l-(4V)/l^2 = 0` `=>4l^3-4V = 0` `=>V = l^3` So, `h = V/l^2 = l^3/l^2 = l` So, all of the sides of the cuboid is `l`. Hence, it is a cube. |
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