Question

For the box to satisfy certain requirements its length must be 3 unit greater than the width.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Width of the box, b = x units

So,

Length of the box, l = x + 3 units

and

Height of the box, h = x - 2 units

$\large\underline{\sf{Solution-46}}$

Now, we have

Width of the box, b = x units

Length of the box, l = x + 3 units

Height of the box, h = x - 2 units

So,

$\rm \: Volume_{(box)} = l \times b \times h$

$\rm \: Volume_{(box)} = (x + 3)\times x \times (x - 2)$

$\rm \: Volume_{(box)} = x(x + 3)(x - 2)$

$\rm \: Volume_{(box)} = x( {x}^{2} - 2x + 3x - 6)$

$\rm \: Volume_{(box)} = x( {x}^{2} + x - 6)$

$\rm\implies \:Volume_{(box)} = {x}^{3} + {x}^{2} - 6x$

So, Option (c) is correct.

$\large\underline{\sf{Solution-47}}$

We have,

Width of the box, b = x units

Length of the box, l = x + 3 units

Height of the box, h = x - 2 units

So, Area of paper required to prepare the box is equal to Total Surface Area of box.

$\rm \: Area\:of\:paper\:required = 2(lb + bh + hl)$

$\rm \: = \: 2[(x + 3)x + x(x - 2) + (x - 2)(x + 3)]$

$\rm \: = \: 2[ {x}^{2} + 3x + {x}^{2} - 2x + {x}^{2} + 3x - 2x - 6]$

$\rm \: = \: 2[ 3{x}^{2} + 2x - 6]$

$\rm \: = \: 6{x}^{2} + 4x - 12 \: sq. \: units$

So, Option (a) is correct.

$\large\underline{\sf{Solution-48}}$

Now, given that

$\rm \: Volume_{(box)} = 18$

$\rm \: {x}^{3} + {x}^{2} - 6x = 18$

$\rm \: {x}^{3} + {x}^{2} - 6x - 18 = 0$

Using, hit and trial method, Let x = 3, we have

$\rm \: = \: {3}^{3} + {3}^{2} - 18 - 18$

$\rm \: = \: 27 + 9 - 36$

$\rm \: = \: 36 - 36$

$\rm \: = \: 0$

So, x - 3 is a factor.

So, by using Long Division Method, we have

$\rm \: (x - 3)( {x}^{2} + 4x + 6) = 0$

$\rm\implies \:x = 3 \: \:$

or

$\rm\implies \: {x}^{2} + 4x + 6 = 0$

Now, Discriminant = 16 - 4 × 1 × 6 = 16 - 24 = - 8 < 0

So, it means equation has no real root.

So,

$\rm\implies \:x = 3$

Hence,

Length of the box, l = 3 + 3 = 6 units

Height of the box, h = 3 - 2 = 1 units

So, option (a) is correct.

$\large\underline{\sf{Solution-49}}$

Now, we have

$\rm \:Area\:of\:paper\:required = \: 6{x}^{2} + 4x - 12 \: sq. \: units$

On substituting x = 3, we get

$\rm \:Area\:of\:paper\:required = \: 6{(3)}^{2} + 4(3) - 12$

$\rm \:Area\:of\:paper\:required = \: 54 + 12 - 12$

$\rm \:Area\:of\:paper\:required = \: 54$

So,

$\rm \:Cost\:of\:paper\: = \: 54 \times 100 = 5400$

So, option (d) is correct.