Question

a square tin sheet of side 12 inches is converted into a box with open top in the following steps: The sheet is placed horizontally. Then, equal sized square, each of side x inches , are cut from the four corners of the sheet. F inally, the four resulting side are bent vertically upwards in the shape of a box. If x is an integer, then what value of x maximize the volume of the box?

Answer 1

Answer:

When we fold the sheet in the form of a box it becomes a cuboid whose length is 12−2x, width is 12−2x and height is x

So volume (V)=l×b×h=(12−2x)

2

×x

∴V=144x−48x

2

+4x

3

For maximum volume of the box,

dx

dV

=0

∴

dx

dV

=144−96x+12x

2

=0

∴(x−2)(x−6)=0

∴x=2orx=6

At x=2, V=[12−2(2)]

2

×2=128........(max)

At x=6, V=[12−2(6)]

2

×6=0

Step-by-step explanation:

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Answer 2

Answer:

2

Step-by-step explanation:

Volume of box = (12 - 2x) * (12 -2x) *  x

= 144x - 48 + 4

differentiate

=> 144 - 96x + 12

solve for x

=> 144 -96x + 12 = 0

=>  - 8x + 12 = 0

=> x = 6, 2

put x in volume

x = 6 , v = 0

x = 2, v = 128 ans