Problem 29 - Precalculus Student Solutions Manual 5th

Question

A can of Blue Runner Red Kidney Beans has surface area \(371 \mathrm{cm}^{2}\). Its height is \(12 \mathrm{cm} .\) What is the radius of the circular top? Round to the nearest hundredth.

Step-by-Step Solution

Answer

The radius of the circular top is approximately 3.75 cm.

Step 1: Understand the shape and formula for surface area of a cylinder

The can is in the shape of a cylinder. The surface area of a cylinder is given by the formula \[ A = 2\pi rh + 2\pi r^2 \] where \( r \) is the radius, \( h \) is the height, and \( A \) is the surface area.

Step 2: Substitute the given values into the formula

Substitute \( A = 371 \mathrm{cm}^2 \) and \( h = 12 \mathrm{cm} \) into the surface area formula. \[ 371 = 2\pi r(12) + 2\pi r^2 \]

Step 3: Simplify the equation

Factor out \( 2\pi r \) from the right side to simplify: \[ 371 = 2\pi r(12 + r) \] which simplifies to \[ 371 = 2\pi r(12 + r) \].

Step 4: Solve for the radius

Divide both sides by \( 2\pi \): \[ \frac{371}{2\pi} = r(12 + r) \]. Use the constant \( \pi \approx 3.14 \), so \[ \frac{371}{2(3.14)} = r(12 + r) \]. Thus, \[ \frac{371}{6.28} = r(12 + r) \]. Perform the division: \[ 59.07 = r(12 + r) \]. Now solve the quadratic equation by reorganizing it as \[ r^2 + 12r - 59.07 = 0 \].

Step 5: Solve the quadratic equation

Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 12 \), and \( c = -59.07 \): \[ r = \frac{-12 \pm \sqrt{12^2 - 4(1)(-59.07)}}{2(1)} \], which simplifies to \[ r = \frac{-12 \pm \sqrt{144 + 236.28}}{2} \], and further to \[ r = \frac{-12 \pm \sqrt{380.28}}{2} \]. Solving inside the square root and then the division gives \[ r = \frac{-12 \pm 19.50}{2} \].

Step 6: Select the positive solution

Select the positive solution since the radius cannot be negative: \[ r = \frac{7.50}{2} = 3.75 \]. Thus, the radius of the circular top of the can is approximately \( 3.75 \mathrm{cm} \).

Key Concepts

cylinder surface area formula

Understanding the surface area of a cylinder is crucial when solving many geometry problems in precalculus. A cylinder is a three-dimensional shape with two parallel circular bases and a curved surface connecting them. The formula to find the surface area of a cylinder is given by \[ A = 2\pi rh + 2\pi r^2 \] where:

• \( A \) is the total surface area
• \( r \) is the radius of the circular base
• \( h \) is the height of the cylinder

. In this formula, \( 2\pi rh \) accounts for the surface area of the cylindrical part (the curved surface), and \( 2\pi r^2 \) accounts for the areas of the two circular bases. Always make sure to use the right values for \( r \) and \( h \), substituting them into the formula for correct results. This becomes especially important when provided with a problem, as you will use these steps to solve for unknown variables.

solving quadratic equations

Solving quadratic equations is a fundamental skill in precalculus and other advanced math courses. A quadratic equation is typically in the form \( ax^2 + bx + c = 0 \) where \( a \), \( b \), and \( c \) are constants. The solutions to these equations can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. This formula might look complicated, but you can break it down into easy steps:

• Identify the coefficients \( a \), \( b \), and \( c \) in your quadratic equation
• Substitute these values into the quadratic formula
• Carry out the calculations inside the square root (\( \sqrt{b^2 - 4ac} \))
• Calculate the two possible solutions by considering both \( + \sqrt{...} \) and \( - \sqrt{...} \)

. When working through problems like determining the radius of a cylinder from its surface area, knowing how to solve a quadratic equation helps you find precise measurements.

geometry in precalculus

Geometry forms a significant part of the precalculus curriculum. It involves understanding shapes, sizes, relative positions, and the properties of space. Geometry will be encountered frequently and can include such topics as:

• Understanding various geometric formulas for areas and volumes of different shapes
• Applying properties and theorems related to angles, lines, and circles
• Solving real-life spatial problems, such as the one with the cylinder’s surface area and radius

. Mastery of geometric principles will not only prepare you for calculus but also for various STEM fields that use these concepts regularly. For instance, in the given problem, knowing how to apply the surface area formula to solve for the radius required an understanding of both algebra and geometry. Thus, practice and familiarity with these concepts will go a long way in your mathematical journey.