Question

Q.26) From the corner of a square card-board of side 12, squares of side x are cut out and the edges are then turned up so as to form an open box, if its volume is
maximum, then x...

Answer 1

Question :- From the corner of a square card-board of side 12, squares of side x are cut out and the edges are then turned up so as to form an open box, if its volume is maximum, then x...? (Excellent Question.)

Solution :-

It has been given that, Squares of side x units are cut from all corners .

Than,

→ Length of card - board left = (12 - 2x)

Similarly,

→ Breadth of card - board left = (12 - 2x)

and,

→ Height of open box formed = side of square cut = x unit.

Than,

→ Volume of box = Length * Breadth * Height

→ V = (12 - 2x) * (12 - 2x) * x

→ V = (12 - 2x)² * x

using (a - b)² = a² + b² - 2ab Now,

→ V = (144 + 4x² - 48x)x

→ V = 4x³ - 48x² + 144x

Now, Differentiating w.r.t x, we get,

→ dV/dx = 12x² - 96x + 144

→ dV/dx = 12(x² - 8x + 12)

Now, for Maximum volume dV/dx is Equal to 0.

So,

→ 12(x² - 8x + 12) = 0

→ x² - 8x + 12 = 0

→ x² - 6x - 2x + 12 = 0

→ x(x - 6) - 2(x - 6) = 0

→ (x - 6)(x - 2) = 0

Putting Both Equal to 0 now, we get,

→ x = 6 and 2.

Again Differentiating, we get ,

→ d²V/dx² = (2x - 8)

Now, if x = 6

→ d²V/dx² = (2*6 - 8) = 12 - 8 = 4 = Positive.

and, if x = 2

→ d²V/dx² = (2*2 - 8) = 4 - 8 = (-4) = Negative.

Therefore, if we put x = 6 we gets minimum volume .

Hence, Volume of open box is Maximum at x = 2.