**An open box with a square base is to be made out of a given quantity of card board of area c ^{2} square units. Show that the maximum volume of the box is c^{3}/ 6**

**√3 cubic units.**

Let *x* be the side of the square base and *y* be the height of the open box.

Given, area of metal sheet = C^{2}

Area of open box = Area of base + area of 4 sides

= x^{2} + 4xy

Now, area of open box is also equal to the area of the metal sheet.

Therefore, C^{2} = x^{2} + 4xy

Now, volume of the box = x^{2}y

On differentiating (1) w.r.t x, we get

Again, differentiating (2) w.r.t. x, we get

Now, for volume to maximum or minimum,

⇒

⇒ C^{2} = 3x^{2}

Or

Now, x is the length of the box, so it can't be negative.

Therefore,

Now, at ,

Therefore, V is maximum.

Hence, maximum value of V is

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