Q. 175.0( 1 Vote )

# An open box with

Answer :

Given,

• The open box has a square base

• The area of the box is c^{2} square units.

• The volume of the box is maximum.

Let us consider,

• The side of the square base of the box be ‘a’ units. (pink coloured in the figure)

• The breadth of the 4 sides of the box will also be ‘a’units (skin coloured part).

• The depth of the box or the length of the sides be ‘h’ units (skin coloured part).

Now, the area of the box =

(area of the base) + 4 (area of each side of the box)

So as area of the box is given c^{2},

c^{2} = a^{2} + 4ah

---- (1)

Let the volume of the newly formed box is :

V = (a)^{2} × (h)

[substituting (1) in the volume formula]

------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with a and then equating it to zero. __This is because if the function f(a) has a maximum/minimum at a point c then f’(c) = 0.__

Differentiating the equation (2) with respect to a:

------- (3)

[Since ]

To find the critical point, we need to equate equation (3) to zero.

c^{2} – 3a^{2} = 0

[as ‘a’ cannot be negative]

__Now to check if this critical point will determine the maximum Volume of the box, we need to check with second differential which needs to be negative.__

Consider differentiating the equation (3) with x:

----- (4)

[Since ]

Now let us find the value of

As , so the function V is maximum at

Now substituting a in equation (1)

Therefore side of wanted box has a base side, is and height of the box, .

Now, the volume of the box is

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