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A box is to be made out of a 8 cm by 20 cm piece of cardboard. Squares of side length x cm will be cut out of each corner, and then the ends and sides will be folded up to form a box with an open top.

(a) Express the volume V of the box as a function of x.

V=

(b) Give the domain of V in interval notation. (Use the fact that length and volume must be positive.)

(c) Find the length L, width W, and height H of the resulting box that maximizes the volume. (Assume that W?L).

L=

W=

H=

(d) The maximum volume of the box is

Respuesta :

Answer:

hi! im sorry if this is wrong, im gonna try my hardest.

V= 160

L= 240

W=34

sry this is prob wrong just wanted to help ;(

MrRoyal

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

The dimensions of the cardboard is give as:

[tex]\mathbf{Length = 8}[/tex]

[tex]\mathbf{Width = 20}[/tex]

Assume the cut-out is x, the dimension of the cardboard would be:

[tex]\mathbf{Length = 8 - 2x}[/tex]

[tex]\mathbf{Width = 20 - 2x}[/tex]

[tex]\mathbf{Height = x}[/tex]

(a) The expression of volume

Volume is calculated as:

[tex]\mathbf{Volume = Length \times Width \times Height}[/tex]

So, we have:

[tex]\mathbf{V = (8 - 2x) \times (20 - 2x) \times x}[/tex]

Hence, the expression for volume is: [tex]\mathbf{V = (8 - 2x) \times (20 - 2x) \times x}[/tex]

(b) The domain of x

Set the volume to 0

[tex]\mathbf{(8 - 2x) \times (20 - 2x) \times x = 0}[/tex]

Split

[tex]\mathbf{8 - 2x = 0\ or \ 20 - 2x = 0 \ x = 0}[/tex]

Solve for x

[tex]\mathbf{x = 4\ or \ x = 10 \ x = 0}[/tex]

x cannot be negative, 0, 4 or greater.

Hence, the domain is: [tex]\mathbf{(0, 4)}[/tex]

(c) The dimension that maximizes the box

We have:

[tex]\mathbf{V = (8 - 2x) \times (20 - 2x) \times x}[/tex]

Expand

[tex]\mathbf{V = 160x - 56x^2 + 4x^3}[/tex]

Differentiate

[tex]\mathbf{V' = 160 - 102x + 12x^2}[/tex]

Set to 0

[tex]\mathbf{160 - 102x + 12x^2 = 0}[/tex]

Using a calculator, we have:'

[tex]\mathbf{x = 2.1, 6.4}[/tex]

Using the domain in (b), we have:

[tex]\mathbf{x = 2.1}[/tex]

Recall that:

[tex]\mathbf{Length = 8 - 2x}[/tex]

[tex]\mathbf{Width = 20 - 2x}[/tex]

[tex]\mathbf{Height = x}[/tex]

So, we have:

[tex]\mathbf{Length = 3.8}[/tex]

[tex]\mathbf{Width = 15.8}[/tex]

[tex]\mathbf{Height = 2.1}[/tex]

The dimension that maximizes the box is 2.8 by 15.8 by 21. cm

(d) The maximum volume

Recall that:

[tex]\mathbf{Volume = Length \times Width \times Height}[/tex]

So, we have:

[tex]\mathbf{Volume = 3.8 \times 15.8 \times 2.1}[/tex]

[tex]\mathbf{Volume = 126.1}[/tex]

Hence, the maximum volume is 126.1 cubic centimeter

Read more about volumes at:

https://brainly.com/question/1578538

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