Calculating Dimensions of a Cylindrical Container with Given Volume

In summary, the question asks to find the dimensions of a cylindrical tennis ball container with a volume of 825π cm3. By setting the volume equation equal to 825π and factoring, we can solve for the radius and height, with the radius being 4 cm and the height being 825/(16π) cm.
  • #1
RohanTalkad
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The question reads, "Find the dimensions of a cylindrical tennis ball container which has the volume of V(x)=8πx3+17πx2+10πx+π such that the volume is exactly 825π cm3. Hint: V = πr2h."

To start off, I set V(x)=825π and moved it to the right side, giving

0 = 8πx3+17πx2+10πx-824π.

Factoring pi, we get 0 = π(x-4)(8x2+49x+206),

Since we can't factor the second bracket, here's where I get confused. My inference is that the radius is 4 cm, and the height is muzzled in that unfactorable bracket. However, having the equation for volume (V = πr2h), I get h = 825/16π.

Can someone verify this for me, please?
 
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  • #2
RohanTalkad said:
The question reads, "Find the dimensions of a cylindrical tennis ball container which has the volume of V(x)=8πx3+17πx2+10πx+π such that the volume is exactly 825 cm3. Hint: V = πr2h."

To start off, I set V(x)=825π and moved it to the right side, giving

0 = 8πx3+17πx2+10πx-824π.

Factoring pi, we get 0 = π(x-4)(8x2+49x+206),

Since we can't factor the second bracket, here's where I get confused. My inference is that the radius is 4 cm, and the height is muzzled in that unfactorable bracket. However, having the equation for volume (V = πr2h), I get h = 825/16π.
Looks fine to me, but you should write that number as 825/(16π). Many people would interpret what you wrote as ##\frac{825}{16}\pi##.
RohanTalkad said:
Can someone verify this for me, please?
 
  • #3
RohanTalkad said:
To start off, I set V(x)=825π
That doesn't sound right. We are given that ##V=825##, not ##V=825\pi##.
 
  • #4
andrewkirk said:
That doesn't sound right. We are given that ##V=825##, not ##V=825\pi##.

Sorry, I forgot to add 825pi as the volume.
 

1. What is a cubic volume problem?

A cubic volume problem is a type of mathematical problem that involves finding the volume of a three-dimensional object. It requires knowledge of the dimensions of the object and the formula for calculating volume.

2. How do I solve a cubic volume problem?

To solve a cubic volume problem, you need to identify the dimensions of the object, such as length, width, and height. Then, use the formula for calculating volume, which is usually V = lwh, where l is the length, w is the width, and h is the height. Plug in the values and solve for the volume.

3. What units are used for cubic volume?

Cubic volume is typically measured in cubic units, such as cubic inches, cubic feet, cubic centimeters, or cubic meters. The unit used will depend on the size of the object being measured.

4. Can I use the same formula for all cubic volume problems?

While the formula for calculating volume (V = lwh) can be used for many cubic volume problems, there are some cases where a different formula may be needed. For example, when finding the volume of a cone or a cylinder, a different formula (V = πr²h) is used.

5. Why is cubic volume important in science?

Cubic volume is important in science because it allows us to measure and compare the amount of space occupied by different objects or substances. It is used in various scientific fields, such as chemistry, physics, and engineering, to calculate the volume of solids, liquids, and gases.

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