Question

A 30.5-ounce can of Hills Bros. \(\begin{array}{llll}\text { coffee } & \text { requires } & 58.9 \pi & \text { square }\end{array}\) inches of aluminum. If its height is 6.4 inches, what is its radius?

Short answer

The radius is 3.1 inches.

Step-by-step solution

- Understand the problem

We need to find the radius of a cylindrical can given its height and surface area. The total surface area provided is 58.9π square inches, and the height is 6.4 inches.

- Recall the formula for the surface area of a cylinder

The total surface area (SA) of a cylinder is given by the formula: \[ SA = 2\text{π}r(h + r) \] where r is the radius and h is the height.

- Substitute the known values into the formula

Substitute the given surface area (58.9π) and height (6.4 inches) into the equation: \[ 58.9\text{π} = 2\text{π}r(6.4 + r) \]

- Simplify the equation

Divide both sides of the equation by π to get rid of π: \[ 58.9 = 2r(6.4 + r) \]

- Expand and solve the quadratic equation

Distribute 2r through (6.4 + r): \[ 58.9 = 12.8r + 2r^2 \]Rearrange the equation to standard quadratic form: \[ 2r^2 + 12.8r - 58.9 = 0 \]

- Solve the quadratic equation

Use the quadratic formula \(r = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a}\), where a = 2, b = 12.8, and c = -58.9: \[ r = \frac{-12.8 \, \pm \, \sqrt{12.8^2 - 4 \, \cdot \, 2 \, \cdot \, -58.9}}{2 \, \cdot \, 2} \]Simplify inside the square root: \[ r = \frac{-12.8 \, \pm \, \sqrt{163.84 + 471.2}}{4} \]\[ r = \frac{-12.8 \, \pm \, \sqrt{635.04}}{4} \]\[ r = \frac{-12.8 \, \pm \, 25.2}{4} \]

- Select the positive solution

The negative radius does not make physical sense, so use the positive solution: \[ r = \frac{-12.8 + 25.2}{4} = \frac{12.4}{4} = 3.1 \]Thus, the radius is 3.1 inches.

Key concepts

cylinder surface area

To solve the problem, we first need to understand the surface area of a cylinder. The surface area is the total area that covers the outside of the cylinder. For a cylinder, the formula is: SA = 2πr(h + r) Here, SA stands for the surface area, r is the radius of the base, and h is the height of the cylinder. This formula includes the area of the two circular bases (2πr^2) and the area of the side (the side forms a rectangle when unrolled, and its area is 2πrh).

quadratic equation

In this exercise, after substituting known values into the surface area formula and simplifying, the problem reduces to a quadratic equation. Understanding quadratics is crucial for solving such problems. The general form of a quadratic equation is: ax^2 + bx + c = 0 Here, x is the variable we are solving for, while a, b, and c are constants. To solve quadratic equations, we can use the quadratic formula: x = ( - b ± √(b² - 4ac) ) / 2a This formula provides solutions for x, which are the points where the quadratic polynomial equals zero.

algebraic substitution

One core skill used in the problem is algebraic substitution. This means inserting known values into an equation. In our case, we replaced the known surface area (58.9π) and height (6.4 inches) into the cylinder surface area formula: 58.9π = 2πr(6.4 + r) By doing so, we convert the problem into a solvable quadratic form. Algebraic substitution simplifies complex equations, enabling step-by-step solution.

radius calculation

Finally, after simplifying the equation and using the quadratic formula, we calculate the radius. After solving: r = ( - 12.8 ± √(12.8² + 4 * 2 * 58.9) ) / 4 Symplifying inside the square root, we choose the positive solution for a physically meaningful radius. Always remember, negative values for geometric measures like length or radius are not practical: r = ( 12.4 / 4 ) = 3.1 inches. Thus, the radius of the cylinder is 3.1 inches.