Question

You are constructing a cardboard box with the dimensions 2 m by 4 m. You then cut equal-size squares ( x by x squares) from each corner so that you may fold the edges to get a box. How long should x be to get a box with the maximum volume?Round your answer to three decimal places

Answer 1

Answer:

x = 0.423 m

Step-by-step explanation:

Volume of the box = length × width × height

The height is given by the length of a side of the square.

height = x

The width and height are gotten by subtracting twice the side of the square form each dimension of the cardboard (since two squares are cut along each dimension).

length = 4 - 2x

width = 2 - 2x

Volume, $V = x(4-2x)(2-2x) = 8x - 12x^2 + 4x^3$

To find the maximum, we differentiate V with respect to x and equate the derivative to 0.

$\dfrac{dV}{dx} = 8-24x+12x^2 = 0$

Simplify by dividing by 4.

$2-6x+3x^2 = 0$

Using the quadratic formula,

x = 0.423 m or x = 1.577 m

To determine which values gives maximum, we differentiate the first derivative which gives

$\dfrac{d^2V}{dx^2} = -6+6x$

The maximum value occurs when $\dfrac{d^2V}{dx^2}$ is negative.

At x = 1.577,

$\dfrac{d^2V}{dx^2} = -6+6(1.577) = 3.462$

This is positive, so it is minimum.

At x = 0.423,

$\dfrac{d^2V}{dx^2} = -6+6(0.423) = -3.462$

This gives a negative value. Therefore, it is a maximum.

Hence, the maximum volume is obtained when x = 0.423 m.

Answer 2

The maximum volume of the box is the highest volume the box can have.

The length of the cut-out x should be 0.423 meter long to get a maximum volume

The dimension of the cardboard is given as:

$\mathbf{Length = 2m}$

$\mathbf{Width = 4m}$

The length of the cut-out is x.

So, the dimension of the box would be

$\mathbf{Length = 2- 2x}$

$\mathbf{Width = 4- 2x}$

$\mathbf{Height =x}$

The volume of the box is then calculated as:

$\mathbf{Volume = Length \times Width \times Height}$

So, we have:

$\mathbf{Volume = (2 - 2x) \times (4 - 2x) \times x}$

Expand

$\mathbf{Volume = 8x -12x^2 + 4x^3}$

Rewrite as:

$\mathbf{V = 8x -12x^2 + 4x^3}$

Differentiate

$\mathbf{V'= 8 -24x + 12x^2}$

Set to 0

$\mathbf{8 -24x + 12x^2 = 0}$

Divide through by 4

$\mathbf{2 -6x + 3x^2 = 0}$

Rewrite as:

$\mathbf{3x^2-6x + 2 = 0}$

Using a calculator, we have:

$\mathbf{x = (0.423, 1.577)}$

The value of x = 1.577 is greater than the dimension of the box.

So, we have:

$\mathbf{x = 0.423}$

Hence, x should be 0.423 meter long

Read more about maximum volumes at:

https://brainly.com/question/17219343