Answer :
To find the radius [tex]\( r \)[/tex] of a circus tent composed of both a cone and a cylinder, given the volume [tex]\( V \)[/tex] and the heights [tex]\( h_1 \)[/tex] and [tex]\( h_2 \)[/tex] of the cone and the cylinder respectively, we start by examining the volume formula:
[tex]\[ V = \frac{1}{3} \pi r^2 h_1 + \pi r^2 h_2 \][/tex]
Our goal is to isolate [tex]\( r \)[/tex] in terms of [tex]\( V \)[/tex], [tex]\( h_1 \)[/tex], and [tex]\( h_2 \)[/tex]. Here’s a step-by-step solution:
1. First, factor out the common term [tex]\( \pi r^2 \)[/tex] from the volume equation:
[tex]\[ V = \pi r^2 \left( \frac{1}{3} h_1 + h_2 \right) \][/tex]
2. Next, divide both sides of the equation by [tex]\( \pi \left( \frac{1}{3} h_1 + h_2 \right) \)[/tex] to isolate [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = \frac{V}{\pi \left( \frac{1}{3} h_1 + h_2 \right)} \][/tex]
3. To further simplify, combine the terms inside the denominator:
[tex]\[ r^2 = \frac{V}{\pi \left( \frac{h_1}{3} + h_2 \right)} \][/tex]
4. Now solve for [tex]\( r \)[/tex] by taking the square root of both sides:
[tex]\[ r = \sqrt{\frac{V}{\pi \left( \frac{h_1}{3} + h_2 \right)}} \][/tex]
5. Simplify the expression slightly for clarity:
[tex]\[ r = \sqrt{\frac{3V}{\pi \left( h_1 + 3h_2 \right)}} \][/tex]
Therefore, the radius [tex]\( r \)[/tex] of the circus tent can be expressed in terms of the volume [tex]\( V \)[/tex], the height of the cone [tex]\( h_1 \)[/tex], and the height of the cylinder [tex]\( h_2 \)[/tex] as follows:
[tex]\[ r = \pm \sqrt{\frac{3V}{\pi \left( h_1 + 3h_2 \right)}} \][/tex]
This solution yields two possible values for [tex]\( r \)[/tex], one positive and one negative. Given that a radius cannot be negative in a real-world geometric context, the practical solution is:
[tex]\[ r = \sqrt{\frac{3V}{\pi \left( h_1 + 3h_2 \right)}} \][/tex]
Thus, this positive expression for [tex]\( r \)[/tex] provides the desired radius of the circus tent.
[tex]\[ V = \frac{1}{3} \pi r^2 h_1 + \pi r^2 h_2 \][/tex]
Our goal is to isolate [tex]\( r \)[/tex] in terms of [tex]\( V \)[/tex], [tex]\( h_1 \)[/tex], and [tex]\( h_2 \)[/tex]. Here’s a step-by-step solution:
1. First, factor out the common term [tex]\( \pi r^2 \)[/tex] from the volume equation:
[tex]\[ V = \pi r^2 \left( \frac{1}{3} h_1 + h_2 \right) \][/tex]
2. Next, divide both sides of the equation by [tex]\( \pi \left( \frac{1}{3} h_1 + h_2 \right) \)[/tex] to isolate [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = \frac{V}{\pi \left( \frac{1}{3} h_1 + h_2 \right)} \][/tex]
3. To further simplify, combine the terms inside the denominator:
[tex]\[ r^2 = \frac{V}{\pi \left( \frac{h_1}{3} + h_2 \right)} \][/tex]
4. Now solve for [tex]\( r \)[/tex] by taking the square root of both sides:
[tex]\[ r = \sqrt{\frac{V}{\pi \left( \frac{h_1}{3} + h_2 \right)}} \][/tex]
5. Simplify the expression slightly for clarity:
[tex]\[ r = \sqrt{\frac{3V}{\pi \left( h_1 + 3h_2 \right)}} \][/tex]
Therefore, the radius [tex]\( r \)[/tex] of the circus tent can be expressed in terms of the volume [tex]\( V \)[/tex], the height of the cone [tex]\( h_1 \)[/tex], and the height of the cylinder [tex]\( h_2 \)[/tex] as follows:
[tex]\[ r = \pm \sqrt{\frac{3V}{\pi \left( h_1 + 3h_2 \right)}} \][/tex]
This solution yields two possible values for [tex]\( r \)[/tex], one positive and one negative. Given that a radius cannot be negative in a real-world geometric context, the practical solution is:
[tex]\[ r = \sqrt{\frac{3V}{\pi \left( h_1 + 3h_2 \right)}} \][/tex]
Thus, this positive expression for [tex]\( r \)[/tex] provides the desired radius of the circus tent.
