Mathematics, 07.04.2020 19:08 aliceohern

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 21 in. long and 5 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.)

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Answer from: loganharper992

The volume of a box is the amount of space in the box.

The dimensions that yield the maximum volume are: 18.60 by 2.60 by 1.20 inches

The dimension of the cardboard is:

$\mathbf{Length = 21}$

$\mathbf{Width = 5}$

Assume the cut-out is x.

So, the dimension of the box becomes

$\mathbf{Length = 21 -2x}$

$\mathbf{Width = 5 -2x}$

$\mathbf{Height = x}$

The volume of the box is:

$\mathbf{V = (21 - 2x) \times (5 - 2x) \times x}$

Open brackets

$\mathbf{V = (21 - 2x) \times (5x - 2x^2)}$

Expand

$\mathbf{V = 105x - 42x^2 -10x^2 + 4x^3}$

$\mathbf{V = 105x -52x^2 + 4x^3}$

Differentiate

$\mathbf{V' = 105 - 104x + 12x^2}$

Set to 0

$\mathbf{105 - 104x + 12x^2 = 0}$

Rewrite as:

$\mathbf{12x^2- 104x +105 = 0}$

Using a calculator, the value of x is:

$\mathbf{x = (7.50, 1.20)}$

When x = 7.50;

$\mathbf{Length = 21 -2(7.50) = 6}$

$\mathbf{Width = 5 -2(7.50) = -10}$

$\mathbf{Height = 7.50}$

The width cannot be negative.

When x = 1.20;

$\mathbf{Length = 21 -2(1.20) = 18.60}$

$\mathbf{Width = 5 -2(1.20) = 2.60}$

$\mathbf{Height = 1.20}$

So, the dimensions that yield the maximum volume are: 18.60 by 2.60 by 1.20 inches

Read more about volumes at:

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Answer from: star030616

The dimension of the box is 18.66 in by 2.66 in by 1.17 in.

Step-by-step explanation:

Given that, The cardboard is 21 in. long and 5 in. wide.

Assume,x be the length of the each sides of the square .

Then the length of the box is = (21-2x) in.

The breadth of the box is =(5-2x) in.

The height of the box is = length of side of the square

= x in.

The volume of the box = length × wide × height

=(21-2x)(5-2x)x

=(105-52x+4x²)x

=( 105 x-52x^2+4x^3 )

Let

V= 105 x-52x^2+4x^3

Differentiating with respect to x

V'=105-104x+12x^2

Again differentiating with respect to x

V''=-104+24x

To find the dimension of the box, we set V'=0

105-104x+12x^2=0

Applying the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , here a=12, b= -104 and c =105

$\Rightarrow x=\frac{-(-104)\pm\sqrt{(-104)^2-4.12.105}}{2.12}$

$=\frac{104\pm 76}{24}$

=7.5, 1.17

For x= 7.5 , the wide of the box will be negative.

∴x=1.17 in.

$V''_{x=1.17}=-104+(24\times 1.17)$

Since at x= 1.17, V''<0 , therefore at x=1.17 in. the volume of the given box will be maximum.

The length of the box is = [21-(1.17×2)]=18.66 in.

The wide of the box is = [5-(1.17×2)]=2.66 in.

The height of the box is = 1.17 in.

The dimension of the box is 18.66 in by 2.66 in by 1.17 in.

Answer from: Quest

The answer of the above is

Answer from: Quest

answer: you need the equation in the question so i can answer it

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