A shipping company needs to determine the number of Styrofoam packing peanuts that fit per square foot in a box that measures 20 ft2.
Question:

1. If 1300 packing peanuts fit in the box, how many packing peanuts are there per square foot in the box?

2. How many packing peanuts fit in a box that is 25% larger in volume?

3. Does the number of packing peanuts per square foot change for the larger box, and why?

Respuesta :

1) There are 65 packing peanuts per square foot

2) 1625 packing peanuts can fit in the new box

3) No, the value of n does not change

Step-by-step explanation:

1)

The total area of the box is

[tex]A=20 ft^2[/tex]

The number of packing peanuts that fit in the box is

N = 1300

Here we are asked to find what is the number of packing peanuts that are per square foot in the box; this can be solved by applying the rule of three:

[tex]\frac{N}{A}=\frac{n}{a}[/tex]

where:

[tex]n[/tex] is the number of packing peanuts that fit in an area of [tex]1ft^2[/tex]

Solving for n,

[tex]n=\frac{Na}{A}=\frac{(1300)(1)}{20}=65[/tex]

So, there are 65 packing peanuts per square foot.

2)

If the area of the box is 25%  larger, than the new area of the box will be:

[tex]A'=A+0.25A[/tex]

And substituting [tex]A=20 ft^2[/tex], we find

[tex]A'=20+0.25\cdot 20=25 ft^2[/tex]

We calculated in part 1) that the number of packing peanuts per square foot is

[tex]n=65[/tex]

Therefore, in order to find how many packing peanuts can fit in the box, we just multiply this number by the new area of the box:

[tex]N=nA'=(65)(25)=1625[/tex]

3)

No, the number of packing peanuts per square foot change for the larger box does not change.

This is because the size of one packing peanuts is always the same, so each packet always occupies the same space in the box, therefore the area of the box increases proportionaly to the number of packets inside.

Mathematically, this can be written as:

[tex]N\propto A[/tex]

Where the constant of proportionality is actually the number of packets that can fit inside one square foot, so the n = 65 that we found in part 1); therefore

[tex]N=nA[/tex]

Learn more about volume and area:

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