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Question
Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is a cube.
Solution
Let x be the side of square base of cuboid and other side be y.
Then the volume of a cuboid with square base,
V = x × x × y
⇒ V = x2y
As the volume of the cuboid is given so volume is taken constantly throughout the question, therefore,
y = `"V"/"x"^2` ....(i)
In order to show that surface area is minimum when the given cuboid is a cube, we have to show S” > 0 and x = y.
Let S be the surface area of cuboid, then
S = x2 + xy + xy + xy + xy + x2
S = 2x2 + 4xy .....(ii)
⇒ S = 2x2 + 4x. `"V"/"x"^2`
⇒ S = 2x2 + `"4V"/"x"` ....(iii)
⇒ `"dS"/"dx" = "4x" - "4V"/"x"^2` ....(iv)
For maximum/minimum value of S, we have `"dS"/"dx" = 0`
⇒ `4"x" - "4V"/"x"^2 = 0 => 4"V" = 4"x"^3`
⇒ V = x3 ....(v)
Putting V = x3 in (i) , we have
y = `"x"^3/"x"^2 = "x"`
Here, y = x ⇒ cuboid is a cube.
Differentiating (iv) w.r.t.x, we have
`("d"^2"S")/"dx"^2 = (4 +(8"V")/"x"^3) >0`
Hence, surface area is minimum when given cuboid is a cube.
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