2. without changing the volume of the can, come up with two new can designs—a taller can and a wider can—and record their dimensions below.
table with columns: height, radius, volume. original: height 12.1 cm, radius 3.1 cm, volume 365.31 cm³. other columns have partial writing (e.g., 2.5 cm radius?)
3. a colleague argues that doubling the radius of the can and halving the height will keep the volume the same. is this true? why or why not?
4. meijer sells bubly water and is offering a president’s day sale on its individual cans. for the price of an original size can of bubly you can choose either:
a. a can with the same height but twice the radius or
b. a can with the same radius but twice the height
which offer is the better deal? explain.
5. suppose you had the option of buying a can with twice the radius and twice the height for triple the price. is this a better or worse deal than either of the two options above? explain.

Question

2. without changing the volume of the can, come up with two new can designs—a taller can and a wider can—and record their dimensions below.
table with columns: height, radius, volume. original: height 12.1 cm, radius 3.1 cm, volume 365.31 cm³. other columns have partial writing (e.g., 2.5 cm radius?)
3. a colleague argues that doubling the radius of the can and halving the height will keep the volume the same. is this true? why or why not?
4. meijer sells bubly water and is offering a president’s day sale on its individual cans. for the price of an original size can of bubly you can choose either:
   a. a can with the same height but twice the radius or
   b. a can with the same radius but twice the height
which offer is the better deal? explain.
5. suppose you had the option of buying a can with twice the radius and twice the height for triple the price. is this a better or worse deal than either of the two options above? explain.

Accepted Answer

### Question 2 Solution:
First, recall the volume formula for a cylinder: $ V=\pi r^{2}h $. The original volume is $ V = \pi\times(3.1)^{2}\times12.1\approx365.31\space cm^{3}$.

#### Taller Can (Decrease Radius, Increase Height):
Let's choose a new radius, say $ r = 2.5\space cm $. We need to find $ h $ such that $ \pi r^{2}h=365.31 $.
## Step 1: Rearrange the volume formula for $ h $
$ h=\frac{V}{\pi r^{2}} $
## Step 2: Substitute $ V = 365.31 $ and $ r = 2.5 $
$ h=\frac{365.31}{\pi\times(2.5)^{2}}\approx\frac{365.31}{19.635}\approx18.6\space cm $

#### Wider Can (Increase Radius, Decrease Height):
Let's choose a new radius, say $ r = 4\space cm $. We need to find $ h $ such that $ \pi r^{2}h = 365.31 $.
## Step 1: Rearrange the volume formula for $ h $
$ h=\frac{V}{\pi r^{2}} $
## Step 2: Substitute $ V = 365.31 $ and $ r = 4 $
$ h=\frac{365.31}{\pi\times(4)^{2}}\approx\frac{365.31}{50.265}\approx7.27\space cm $

Filling the table:
| Height | 12.1 cm | 18.6 cm | 7.27 cm |
|--------|---------|---------|---------|
| Radius | 3.1 cm  | 2.5 cm  | 4 cm    |
| Volume | $ 365.31\space cm^{3} $ | $ 365.31\space cm^{3} $ | $ 365.31\space cm^{3} $ |

### Question 3 Solution:
The volume of a cylinder is given by $ V=\pi r^{2}h $. Let the original radius be $ r $ and original height be $ h $, so original volume $ V_{1}=\pi r^{2}h $.

If we double the radius ($ r_{2} = 2r $) and halve the height ($ h_{2}=\frac{h}{2} $), the new volume $ V_{2}=\pi(2r)^{2}\times\frac{h}{2} $.

## Step 1: Simplify $ V_{2} $
$ V_{2}=\pi\times4r^{2}\times\frac{h}{2}=2\pi r^{2}h $
## Step 2: Compare $ V_{1} $ and $ V_{2} $
We can see that $ V_{2} = 2V_{1} $, which means the volume doubles, not stays the same. So the colleague's argument is false because doubling the radius (which is squared in the volume formula) has a larger impact than halving the height.

# Answer (Question 3):
No, it is not true. When the radius is doubled ($ r
ightarrow2r $) and height is halved ($ h
ightarrow\frac{h}{2} $), the new volume $ V_{new}=\pi(2r)^{2}(\frac{h}{2}) = 2\pi r^{2}h=2V_{original} $, so the volume doubles, not remains the same.

### Question 4 Solution:
Let the original radius be $ r $ and original height be $ h $, so original volume $ V=\pi r^{2}h $.

#### Option a: Same height, twice the radius ($ r_{a}=2r $, $ h_{a}=h $)
Volume of option a: $ V_{a}=\pi(2r)^{2}h=\pi\times4r^{2}h = 4\pi r^{2}h=4V $

#### Option b: Same radius, twice the height ($ r_{b}=r $, $ h_{b}=2h $)
Volume of option b: $ V_{b}=\pi r^{2}(2h)=2\pi r^{2}h = 2V $

Since $ V_{a}=4V $ and $ V_{b}=2V $, and both cost the same as the original can (which has volume $ V $), option a gives a larger volume (4 times the original) compared to option b (2 times the original). So option a is a better deal.

# Answer (Question 4):
Option a (a can with the same height but twice the radius) is the better deal. The volume of a cylinder is $ V = \pi r^{2}h $. For option a, with $ r
ightarrow2r $ and $ h $ constant, the volume becomes $ \pi(2r)^{2}h=4\pi r^{2}h = 4V $ (where $ V $ is the original volume). For option b, with $ h
ightarrow2h $ and $ r $ constant, the volume becomes $ \pi r^{2}(2h)=2\pi r^{2}h = 2V $. Since both cost the same as the original can, option a gives 4 times the original volume while option b gives 2 times, so option a is better.

### Question 5 Solution:
Let the original radius be $ r $ and original height be $ h $, so original volume $ V=\pi r^{2}h $.

If the new can has twice the radius ($ r_{new}=2r $) and twice the height ($ h_{new}=2h $), its volume $ V_{new}=\pi(2r)^{2}(2h) $.

## Step 1: Simplify $ V_{new} $
$ V_{new}=\pi\times4r^{2}\times2h = 8\pi r^{2}h=8V $

The original cost is for volume $ V $. Option a (from question 4) gives $ 4V $ for the same price as $ V $, and option b gives $ 2V $ for the same price as $ V $. The new option costs triple the price of the original. Let's compare the volume per unit price:

- Original: $ \frac{V}{P} $ (where $ P $ is original price)
- Option a: $ \frac{4V}{P} $
- Option b: $ \frac{2V}{P} $
- New option: $ \frac{8V}{3P}\approx2.67\frac{V}{P} $

Since $ 4\frac{V}{P}>2.67\frac{V}{P}>2\frac{V}{P} $, the new option (triple price for $ 8V $) is worse than option a (same price as original for $ 4V $) and better than option b (same price as original for $ 2V $)? Wait, no, let's re - calculate:

Wait, the original price is for $ V $. The new option is triple the price ($ 3P $) for $ 8V $. So the volume per price for new option is $ \frac{8V}{3P}\approx2.67\frac{V}{P} $. Option a is $ \frac{4V}{P} $, option b is $ \frac{2V}{P} $. So $ \frac{4V}{P}>\frac{8V}{3P}>\frac{2V}{P} $. So the new option is a worse deal than option a (since you get less volume per price) but a better deal than option b (since you get more volume per price than option b). But generally, compared to the original "price for $ V $", the new option: for $ 3P $ you get $ 8V $, so per $ P $ you get $ \frac{8}{3}V\approx2.67V $, while option a gives $ 4V $ per $ P $, so it's worse than option a.

# Answer (Question 5):
This is a worse deal than option a (same height, twice radius) and a better deal than option b (same radius, twice height). The volume of the new can is $ V_{new}=\pi(2r)^{2}(2h)=8\pi r^{2}h = 8V $ (where $ V $ is original volume). The original price is for $ V $, option a (same price as original) gives $ 4V $, option b (same price as original) gives $ 2V $, and the new option costs $ 3P $ (triple original price) for $ 8V $. The volume per unit price for the new option is $ \frac{8V}{3P}\approx2.67\frac{V}{P} $, for option a it's $ \frac{4V}{P} $, and for option b it's $ \frac{2V}{P} $. Since $ 4\frac{V}{P}>\frac{8}{3}\frac{V}{P}>2\frac{V}{P} $, the new option is worse than option a (less volume per price) but better than option b (more volume per price).

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QUESTION IMAGE

Question

Explanation:

Question 2 Solution:

Taller Can (Decrease Radius, Increase Height):

Step 1: Rearrange the volume formula for \( h \)

Step 2: Substitute \( V = 365.31 \) and \( r = 2.5 \)

Wider Can (Increase Radius, Decrease Height):

Step 1: Rearrange the volume formula for \( h \)

Step 2: Substitute \( V = 365.31 \) and \( r = 4 \)

Question 3 Solution:

Step 1: Simplify \( V_{2} \)

Step 2: Compare \( V_{1} \) and \( V_{2} \)

Answer:

Question 4 Solution:

Option a: Same height, twice the radius (\( r_{a}=2r \), \( h_{a}=h \))

Option b: Same radius, twice the height (\( r_{b}=r \), \( h_{b}=2h \))