# You have a open box that is made from a 16 in. x30 in. piece of cardboard. When you cut out the squares of equal size from the 4 corners and bending it. What size should the squares be to get this box to work with the largest volume?

##### 1 Answer
Mar 4, 2018

$3 \frac{1}{3}$ inches to be cut from $4$ corners and bend to get
box for maximum volume of
$725.93$ cubic inches.

#### Explanation:

Card board size is $L = 30 \mathmr{and} W = 16$ inches

Let $x$ in square is cut from $4$ corners and bended into

a box whos size is now $L = 30 - 2 x , W = 16 - 2 x \mathmr{and} h = x$

inches. Volume of the box is $V = \left(30 - 2 x\right) \left(16 - 2 x\right) x$ cubic

inches. $V = \left(4 {x}^{2} - 92 x + 480\right) x = 4 {x}^{3} - 92 {x}^{2} + 480 x$.

For maximum value $\frac{\mathrm{dV}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dV}}{\mathrm{dx}} = 12 {x}^{2} - 184 x + 480 = 12 \left({x}^{2} - \frac{46}{3} x + 40\right)$

$12 \left({x}^{2} - 12 x - \frac{10}{3} x + 40\right) = 12 \left(x \left(x - 12\right) - \frac{10}{3} \left(x - 12\right)\right)$

or $12 \left(x - 12\right) \left(x - \frac{10}{3}\right) = 0 \therefore$ Critical points are

x=12 ,x=10/3; x !=12 , as $24$ inches cannot be removed from

$16$ inches width. So $x = \frac{10}{3} \mathmr{and} 3 \frac{1}{3}$ inches to be cut.

Slope test may be examined at$\left(x = 3 \mathmr{and} x = 4\right)$ to show

volume is maximum. $\frac{\mathrm{dV}}{\mathrm{dx}} = 12 \left(x - 12\right) \left(x - \frac{10}{3}\right)$

$\frac{\mathrm{dV}}{\mathrm{dx}} \left(3\right) = \left(+\right) \mathmr{and} \frac{\mathrm{dV}}{\mathrm{dx}} \left(4\right) = \left(-\right)$. Slope at critical point

is from positive to negative , so the volume is maximum.

The maximum volume is $V = \left(30 - \frac{20}{3}\right) \left(16 - \frac{20}{3}\right) \frac{10}{3}$or

$V = \left(30 - \frac{20}{3}\right) \left(16 - \frac{20}{3}\right) \frac{10}{3} \approx 725.93$ cubic inches. [Ans]