Mathematics, 01.07.2019 02:30 coolgirl5679

On a rectangular piece of cardboard with perimeter 10 inches, three parallel and equally spaced creases are made (see figure 1). the cardboard is then folded along the creases to make a rectangular box with open ends (see figure 2). letting x represent the distance (in inches) between the creases, use the aleks graphing calculator to find the value of x that maximizes the volume enclosed by this box. then give the maximum volume. round your responses to two decimal places.

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Answer from: cocobelle

The volume of a shape is the amount of space in it.

The value of x that maximizes the volume is: 0.83The maximum volume is 1.16

The given parameter is:

$\mathbf{Perimeter = 10}$

Let the height of the cardboard be h.

So, the perimeter of the cardboard (see attachment) is:

$\mathbf{Perimeter = x + x + x + x+x + x + x + x + h + h}$

$\mathbf{Perimeter = 8x + 2h}$

Substitute $\mathbf{Perimeter = 10}$

$\mathbf{8x + 2h = 10}$

Divide through by 2

$\mathbf{4x + h = 5}$

Make h the subject

$\mathbf{h = 5 -4x}$

The volume of the cardboard is then calculated as:

$\mathbf{V = x^2h}$

Substitute $\mathbf{h = 5 -4x}$

$\mathbf{V = x^2(5 - 4x)}$

Expand

$\mathbf{V = 5x^2- 4x^3}$

Differentiate

$\mathbf{V' = 10x- 12x^2}$

Set to 0

$\mathbf{10x- 12x^2 = 0}$

Rewrite as:

$\mathbf{- 12x^2 = -10x}$

Divide through by -x

$\mathbf{12x = 10}$

Divide through by 12

$\mathbf{x = \frac{10}{12}}$

$\mathbf{x = \frac{5}{6}}$

$\mathbf{x = 0.83}$

Recall that:

$\mathbf{V = x^2(5 - 4x)}$

So, we have:

$\mathbf{V = (\frac 56)^2 \times (5 - 4 \times \frac 56)}$

$\mathbf{V = \frac{25}{36} \times (5 - \frac{10}{3})}$

Take LCM

$\mathbf{V = \frac{25}{36} \times \frac{15 - 10}{3}}$

$\mathbf{V = \frac{25}{36} \times \frac{5}{3}}$

$\mathbf{V = \frac{125}{108}}$

$\mathbf{V = 1.16}$

Hence,

The value of x that maximizes the volume is: 0.83The maximum volume is 1.16

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