# Question #6d4ae

Oct 23, 2016

$\frac{2}{3} \text{ inch}$

#### Explanation:

Let the size of square to be cut from each corner $= x \text{ inch}$
Volume of the open box so formed $V = l \times w \times h = \left(8 - 2 x\right) \times \left(3 - 2 x\right) \times x$
$\implies V = \left(24 - 22 x + 4 {x}^{2}\right) \times x$
$\implies V = 24 x - 22 {x}^{2} + 4 {x}^{3}$
To find maximum value we need to differentiate $V$ with respect to $x$ and equate it to $0$
$\frac{\mathrm{dV}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(24 x - 22 {x}^{2} + 4 {x}^{3}\right)$, setting it equal to zero we get
$24 - 44 x + 12 {x}^{2} = 0$, dividing both sides by $4$ and rearranging we get
$3 {x}^{2} - 11 x + 6 = 0$
Using split the middle term method to find the roots of above we get
$3 {x}^{2} - 9 x - 2 x + 6 = 0$
$\implies 3 x \left(x - 3\right) - 2 \left(x - 3\right) = 0$
$\implies \left(x - 3\right) \left(3 x - 2\right) = 0$
We get the roots as
$x = 3 , \frac{2}{3}$

For $x = 3$, we see that width $w$ of the box becomes negative. Hence, ignoring this root we have size of the square $\frac{2}{3} \text{ inch}$