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To determine the radius of the circular top of the can of beans given that the surface area is 394 cm² and the height is 18 cm, we need to use the formula for the surface area of a cylinder. The surface area [tex]\(A\)[/tex] of a cylinder is given by the formula:
[tex]\[ A = 2\pi r^2 + 2\pi rh \][/tex]
Where:
- [tex]\(r\)[/tex] is the radius of the base,
- [tex]\(h\)[/tex] is the height of the cylinder,
- and [tex]\(\pi\)[/tex] is the mathematical constant approximately equal to 3.14159.
Given:
[tex]\[ A = 394 \, \text{cm}^2 \][/tex]
[tex]\[ h = 18 \, \text{cm} \][/tex]
Plugging the given values into the formula, we have:
[tex]\[ 394 = 2\pi r^2 + 2\pi r \cdot 18 \][/tex]
Simplify the equation:
[tex]\[ 394 = 2\pi r^2 + 36\pi r \][/tex]
To solve for [tex]\(r\)[/tex], we can rearrange the equation into standard quadratic form:
[tex]\[ 2\pi r^2 + 36\pi r - 394 = 0 \][/tex]
This is a quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], where:
[tex]\[ a = 2\pi \][/tex]
[tex]\[ b = 36\pi \][/tex]
[tex]\[ c = -394 \][/tex]
We solve this quadratic equation using the quadratic formula:
[tex]\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values for [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ r = \frac{-36 \pi \pm \sqrt{(36\pi)^2 - 4 \cdot 2\pi \cdot (-394)}}{2 \cdot 2\pi} \][/tex]
[tex]\[ r = \frac{-36\pi \pm \sqrt{1296\pi^2 + 3152\pi}}{4\pi} \][/tex]
[tex]\[ r \approx \frac{-36\pi \pm \sqrt{(1296\pi^2 + 3152\pi)}}{4\pi} \][/tex]
From the solutions of this quadratic equation, the positive root represents the physical radius of the cylinder. Upon solving, we obtain two roots:
[tex]\[ r_1 \approx 2.99 \][/tex]
[tex]\[ r_2 \approx - \text{(negative value, not physically meaningful)} \][/tex]
Since the radius must be a positive value, we take the positive solution:
[tex]\[ r \approx 2.99 \][/tex]
Thus, the radius of the circular top is approximately [tex]\(2.99\)[/tex] cm when rounded to the nearest hundredth.
[tex]\[ A = 2\pi r^2 + 2\pi rh \][/tex]
Where:
- [tex]\(r\)[/tex] is the radius of the base,
- [tex]\(h\)[/tex] is the height of the cylinder,
- and [tex]\(\pi\)[/tex] is the mathematical constant approximately equal to 3.14159.
Given:
[tex]\[ A = 394 \, \text{cm}^2 \][/tex]
[tex]\[ h = 18 \, \text{cm} \][/tex]
Plugging the given values into the formula, we have:
[tex]\[ 394 = 2\pi r^2 + 2\pi r \cdot 18 \][/tex]
Simplify the equation:
[tex]\[ 394 = 2\pi r^2 + 36\pi r \][/tex]
To solve for [tex]\(r\)[/tex], we can rearrange the equation into standard quadratic form:
[tex]\[ 2\pi r^2 + 36\pi r - 394 = 0 \][/tex]
This is a quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex], where:
[tex]\[ a = 2\pi \][/tex]
[tex]\[ b = 36\pi \][/tex]
[tex]\[ c = -394 \][/tex]
We solve this quadratic equation using the quadratic formula:
[tex]\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values for [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ r = \frac{-36 \pi \pm \sqrt{(36\pi)^2 - 4 \cdot 2\pi \cdot (-394)}}{2 \cdot 2\pi} \][/tex]
[tex]\[ r = \frac{-36\pi \pm \sqrt{1296\pi^2 + 3152\pi}}{4\pi} \][/tex]
[tex]\[ r \approx \frac{-36\pi \pm \sqrt{(1296\pi^2 + 3152\pi)}}{4\pi} \][/tex]
From the solutions of this quadratic equation, the positive root represents the physical radius of the cylinder. Upon solving, we obtain two roots:
[tex]\[ r_1 \approx 2.99 \][/tex]
[tex]\[ r_2 \approx - \text{(negative value, not physically meaningful)} \][/tex]
Since the radius must be a positive value, we take the positive solution:
[tex]\[ r \approx 2.99 \][/tex]
Thus, the radius of the circular top is approximately [tex]\(2.99\)[/tex] cm when rounded to the nearest hundredth.
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