# Math Help - Finding the contact surface area of a plastic bag

1. ## Finding the contact surface area of a plastic bag

Hello,

A flat sealed rectangular plastic bag (ex. ziploc bag) can hold max 50 mL water and has:
L = 14.3 cm
W = 7.5 cm

If the bag was fill with 10 mL of water then what is the surface area where the bag contact with water (the bag was hang up-right position).

2. Originally Posted by eco1
Hello,

A plastic bag (ex. ziploc bag) can hold max 50 mL water and has:
L = 14.3 cm
W = 7.5 cm

If the bag was fill with 10 mL of water then what is the surface area where the bag contact with water (the bag was hang up-right position).
First here is my disclaimer. I think this question is worded very poorly and the answer is ambiguous.

First you will need to know that 1mL=1cc (cubic centimter)

Now here is the first problem we don't know the third dimension of the bag or its true shape.

Assumption 1: It is a rectangular solid. If that is the case then its height can be found via the formula

$V=lwh \iff 50\text{cm}^3=(14.3c.m)(7.5cm)h \iff h=\frac{200}{429}cm$

Now here is problem number 2: The answer depends on which side is the bottom.

Again we use the volume formula but this time the volume of fluid 10mL.

$10\text{cm}^3=lwh$

Now we can pick two of the variables l,w, or h and plug them in to solve for the third. For example lets use l and w.

$10\text{cm}^3=(14.3\text{cm}^2)(7.5\text{cm}^3)h \iff h=\frac{40}{429}$

So now the water will cover the bottom and up to h on the 4 other sides and will not reach the top so the surface area is

$S=(14.3\text{cm})(7.5\text{cm})+2\left( \frac{40\text{cm}}{429}\right)(14.3\text{cm})+2\left( \frac{40\text{cm}}{429}\right)(7.5\text{cm})=\frac{253801}{17160}\text{cm}^2 \approx 14.8\text{cm}^2$

If you repeat the above with l and h or h and w you will get two different answers!

3. Thanks TheEmptySet.
If we filled the ziploc bag with water, then it will look like a pillow. Should we apply the circumference of bag (2W) and treat the shape of the bag with water like a cylinder with flattened ends?

4. At this point, if time permits, I would ask for clarification of the question. If that is not the case then your guess is as good as anyone's what the true shape of the ziplock bag is.

5. could you please let me know which part is unclear? thanks.

I'm thinking, the L will change based on the volume of water inside the bag. So we must know the depth of 10mL then x 2 sides of contact surface area?

6. The problem itself is unclear because it does not say which side is "up" and which surface you are to find the area of. You can do this: you are given that two dimensions of the sack are 14.3 cm and 7.5 cm. Taking "x" to be the length of the third dimension in cm, we have volume= (14.3)(7.5)x= 107.25x= 50 cubic cm (1 milliliter is 1 cubic cm) so that x= 10/107.25 cm.

Now, the problem is which of the two given dimensions is also in contact with the ground? If it is the 14.3 cm side, then the area in contact with the ground is (14.3)(50/107.25) square cm. Your description makes it sound like the longest side is vertical so the area in contact with the ground is 7.5 cm by (50/107.25) cm and so would have area (7.5)(50/107.25) cm.

7. Thank you HallsofIvy. So the formula for the area in contact is: W x (V/Area) correct? Could you please explain how you got that equation?