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To solve for the radius of the circular top of the can with a given surface area of 323 cm² and a height of 16 cm, we follow these steps:
1. Understand the Given Information:
- Surface area (S) = 323 cm²
- Height (h) = 16 cm
2. Formula for the Surface Area of a Cylinder:
The surface area of a cylinder includes the area of the top and bottom (two circles) and the lateral surface area (the side of the cylinder).
[tex]\[ S = 2\pi r^2 + 2\pi rh \][/tex]
where [tex]\( r \)[/tex] is the radius.
3. Rearrange the Formula:
Group the terms involving [tex]\( r \)[/tex]:
[tex]\[ S = 2\pi r (r + h) \][/tex]
Substitute the given values:
[tex]\[ 323 = 2\pi r (r + 16) \][/tex]
4. Simplify and Solve for [tex]\( r \)[/tex]:
We need to solve for [tex]\( r \)[/tex] in the equation:
[tex]\[ 323 = 2\pi r (r + 16) \][/tex]
Divide both sides by [tex]\( 2\pi \)[/tex]:
[tex]\[ \frac{323}{2\pi} = r(r + 16) \][/tex]
Simplify the left-hand side:
[tex]\[ \frac{323}{2\pi} \approx 51.39 \][/tex]
So, we have:
[tex]\[ r^2 + 16r = 51.39 \][/tex]
5. Form a Quadratic Equation:
Write the equation in standard quadratic form:
[tex]\[ r^2 + 16r - 51.39 = 0 \][/tex]
6. Solve the Quadratic Equation:
Use the quadratic formula [tex]\( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = 16 \)[/tex], and [tex]\( c = -51.39 \)[/tex].
[tex]\[ r = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot (-51.39)}}{2 \cdot 1} \][/tex]
Calculate the discriminant:
[tex]\[ \text{Discriminant} = 16^2 - 4 \cdot 1 \cdot (-51.39) = 256 + 4 \cdot 51.39 \][/tex]
[tex]\[ \text{Discriminant} \approx 455.56 \][/tex]
Take the square root of the discriminant:
[tex]\[ \sqrt{455.56} \approx 21.35 \][/tex]
7. Find the Roots:
Substitute back into the quadratic formula:
[tex]\[ r = \frac{-16 \pm 21.35}{2} \][/tex]
This gives us two solutions:
[tex]\[ r = \frac{5.35}{2} \approx 2.68 \][/tex]
[tex]\[ r = \frac{-37.35}{2} \approx -18.68 \][/tex]
Since the radius cannot be negative, we consider the positive root.
8. Round the Final Answer:
Round the positive root to the nearest hundredth:
[tex]\[ r \approx 2.68 \][/tex]
Therefore, the radius of the circular top of the can is approximately 2.68 cm.
1. Understand the Given Information:
- Surface area (S) = 323 cm²
- Height (h) = 16 cm
2. Formula for the Surface Area of a Cylinder:
The surface area of a cylinder includes the area of the top and bottom (two circles) and the lateral surface area (the side of the cylinder).
[tex]\[ S = 2\pi r^2 + 2\pi rh \][/tex]
where [tex]\( r \)[/tex] is the radius.
3. Rearrange the Formula:
Group the terms involving [tex]\( r \)[/tex]:
[tex]\[ S = 2\pi r (r + h) \][/tex]
Substitute the given values:
[tex]\[ 323 = 2\pi r (r + 16) \][/tex]
4. Simplify and Solve for [tex]\( r \)[/tex]:
We need to solve for [tex]\( r \)[/tex] in the equation:
[tex]\[ 323 = 2\pi r (r + 16) \][/tex]
Divide both sides by [tex]\( 2\pi \)[/tex]:
[tex]\[ \frac{323}{2\pi} = r(r + 16) \][/tex]
Simplify the left-hand side:
[tex]\[ \frac{323}{2\pi} \approx 51.39 \][/tex]
So, we have:
[tex]\[ r^2 + 16r = 51.39 \][/tex]
5. Form a Quadratic Equation:
Write the equation in standard quadratic form:
[tex]\[ r^2 + 16r - 51.39 = 0 \][/tex]
6. Solve the Quadratic Equation:
Use the quadratic formula [tex]\( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = 16 \)[/tex], and [tex]\( c = -51.39 \)[/tex].
[tex]\[ r = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot (-51.39)}}{2 \cdot 1} \][/tex]
Calculate the discriminant:
[tex]\[ \text{Discriminant} = 16^2 - 4 \cdot 1 \cdot (-51.39) = 256 + 4 \cdot 51.39 \][/tex]
[tex]\[ \text{Discriminant} \approx 455.56 \][/tex]
Take the square root of the discriminant:
[tex]\[ \sqrt{455.56} \approx 21.35 \][/tex]
7. Find the Roots:
Substitute back into the quadratic formula:
[tex]\[ r = \frac{-16 \pm 21.35}{2} \][/tex]
This gives us two solutions:
[tex]\[ r = \frac{5.35}{2} \approx 2.68 \][/tex]
[tex]\[ r = \frac{-37.35}{2} \approx -18.68 \][/tex]
Since the radius cannot be negative, we consider the positive root.
8. Round the Final Answer:
Round the positive root to the nearest hundredth:
[tex]\[ r \approx 2.68 \][/tex]
Therefore, the radius of the circular top of the can is approximately 2.68 cm.
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