Question

Packaging By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. If the cardboard is 15 in. long and 7 in. wide, find the dimensions (in inches) of the box that will yield the maximum volume. (Round your answers to two decimal places if necessary.)

Answer 1

Answer:

Height: 3/2 inches

Length:  12 inches

Width: 4 inches

Step-by-step explanation:

Let x is the side length of the square

The height of the box by cutting squares off :x

• The new length of the cardboard = 15 -2x (because we cut from 4 corners)
• The new width of the cardboard = 7 -2x (because we cut from 4 corners)

The new volume of it is:

V = (15 -2x) (7 -2x) x

<=> V = $4x^3 -44x^2 +105x$

To maximum volume, we use the first derivative of the volume

$\dfrac{dV}{dx} = 0$

<=> $12x^2-88x+105=0$

<=> $\left(2x-3\right)\left(6x-35\right) = 0$

<=> 2x -3 = 0 or 6x -35 = 0

<=> x = 3/2 or x = 35/6

To determine which value of x gives a maximum, we evaluate

$\dfrac{d^2V}{dx^2}$ = 24x -88

• If  x = 3/2, we have:

$\dfrac{d^2V}{dx^2}$= 24(3/2) -88  = -52

• If x = 35/6, we have:

$\dfrac{d^2V}{dx^2}$ = 24(35/6) -88 = 52

We choose x = 3/2 to have the maximum volume because the value of x that gives a negative value is maximum.

So the dimensions (in inches) of the box is:

Height: 3/2 inches

Length: 15-2(3/2) = 12 inches

Width: 7 - 2(3/2) = 4 inches