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What should be the side of the squares which has to be cut out from the corners of square sheet of side 10 cm, so that the volume of the open box formed by the sheet should be maximum ?
Given :
Side of the square sheet $= 10 cm$.
Four squares cut out from the corners of the square sheet.
To find :
We have to find the side of the squares to be cut out, to get the maximum volume of cuboid which is formed by the sheet.
Solution :
Let the side of square to be cut out be 'x' cm.
After cutting out the remaining side of square sheet $=10 - x - x = 10 - 2x $.
From the figure,
Length of the cuboid $= 10 - 2x $
Breadth of the cuboid $= 10 - 2x $
Height of the cuboid $= x $
We need to find, for what value of 'x' the volume of cuboid will be maximum.
Volume of cuboid $= Length \times Breadth \times Height $
Volume of cuboid $= (10 - 2x) \times (10 - 2x) \times x $
To form the cuboid from the given sheet. we can cut out the squares of side
1 cm, 2 cm, 3 cm, 4 cm.
When $x = 1$
Volume of cuboid $= (10 - 2(1)) \times (10 - 2(1)) \times 1 $
$ = (10 - 2) \times (10 - 2) \times 1 $
$= 8 \times 8 \times 1 $
Volume of cuboid $= 64 cm^3$.
When $x = 2$
Volume of cuboid $= (10 - 2(2)) \times (10 - 2(2)) \times 2 $
$ = (10 - 4) \times (10 - 4) \times 2 $
$= 6 \times 6 \times 2 $
Volume of cuboid $= 72 cm^3$.
When $x = 3$
Volume of cuboid $= (10 - 2(3)) \times (10 - 2(30)) \times 3 $
$ = (10 - 6) \times (10 - 6) \times 3 $
$= 4 \times 4 \times 3 $
Volume of cuboid $= 48 cm^3$.
When $x = 4$
Volume of cuboid $= (10 - 2(4)) \times (10 - 2(4)) \times 4 $
$ = (10 - 8) \times (10 - 8) \times 4 $
$= 2 \times 2 \times 4 $
Volume of cuboid $= 16 cm^3$.
So, when $x = 2$ , the volume of cuboid $= 72 cm^3$. It is the maximum volume.
Therefore, to get the maximum volume of cuboid, the side of squares to be cut out is 2 cm.
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