Question

03] Find the value of largest boxes that can be
made by cutting equal square out of curner of a
pies of cardbord of diamension 15cm by 24cm & they
tremine up the side​

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

●Let the side of the square to be cut off be 'x' cm.

●Now, flaps are folded to make an open box having dimensions

•Length of box = 24 - 2x cm

•Breadth of box = 15 - 2x cm

•Height of box = 'x' cm

☆ Volume of box (V) is given by

$:  \implies  \tt \: V \: = \: length \times breadth \times height$

$:  \implies  \tt \: V \: = (24 - 2x)(15 - 2x)x$

$:  \implies  \tt \: V \: = \: (360 - 48x - 30x + 4 {x}^{2} ) \times x$

$:  \implies  \tt \: V = ( 4{x}^{2} - 78x + 360) \times x$

$:  \implies  \tt \: V = {4x}^{ 3 } - 78 {x}^{2} + 360x$

☆ On differentiating w. r. t. x both sides, we get

$:  \implies  \tt \: \dfrac{dV}{dx} = \dfrac{d}{dx} ( {4x}^{3} - 78 {x}^{2} + 360x)$

$:  \implies  \tt \: \dfrac{dV}{dx} = \: {12x}^{2} - 156x + 360$

☆ For maximum or minimum value,

$:  \implies  \tt \: \dfrac{dV}{dx} = \: 0$

$:  \implies  \tt \: {12x}^{2} - 156x + 360 = 0$

$:  \implies  \tt \:12( {x}^{2} - 13x + 30) = 0$

$:  \implies  \tt \: {x}^{2} - 13x + 30 = 0$

$:  \implies  \tt \: {x}^{2} - 10x - 3x + 30 = 0$

$:  \implies  \tt \:x(x - 10) - 3(x - 10) = 0$

$:  \implies  \tt \:(x - 3)(x - 10) = 0$

$:  \implies  \tt \:x = 3 \: \: or \: \: x = 10$

• But x = 10, rejected because otherwise breadth is - 5 units which in absurd.

• Now, Consider

$:  \implies  \tt \: \dfrac{ {d}^{2} V}{d {x}^{2} } =6x - 156$

• When x = 3, we get

$:  \implies  \tt \: \dfrac{ {d}^{2} V}{d {x}^{2} } = \: 6 \times 3 \: - 156$

$:  \implies  \tt \: \dfrac{ {d}^{2} V}{d {x}^{2} } = \: 18 - 156$

$:  \implies  \tt \: \dfrac{ {d}^{2} V}{d {x}^{2} } = \: - 138$

$:  \implies  \tt \: \dfrac{ {d}^{2} V}{d {x}^{2} } < 0$

$:  \implies  \tt \:V \: is \: maximum \: when \: x \: = 3$

• Dimension of box

$\tt \: Length \: of \: box = 24 - 2x = 24 - 6 = 18 \: cm$

$\tt \:Breadth \: of \: box = 15 - 2x = 15 - 6 = 9 \: cm$

$\tt \:Height of box = x \: = \: 3 \: cm$

$\boxed{ \red{\tt \:Volume \: of \: box = 18 \times 9 \times 3 = 486 \: {cm}^{3} }}$