Exercise 56 Page 212

Practice makes perfect

a To solve this question we will need the formula for the area of a circle in terms of its diameter.

A=π d^2/4 ⇔ A=π/4d^2 We are given both an upper and lower bound for the cross sectional area, which is the area of a circle with diameter x. Let's write inequalities representing these constraints.

Information	Inequality
The cross sectional area must be at least 35 square inches.	π/4x^2≥ 35
The cross sectional area should not be more than 42 square inches.	π/4x^2≤ 42

Let's graph these inequalities one at a time. We need the graph of the quadratic representing the area. Since x is a length, we only draw the graph for positive x-values. The solution to the inequality expressing the lower bound is the set of all x for which the area graph is above the line y=35.

The solution to the inequality expressing the upper bound is the set of all x for which the area graph is below the line y=42.

The solution set of the two inequalities together is the intersection of the two individual solution sets. It is the set of all x for which the graph is between the two horizontal lines.

b We can use a calculator to find the intersection points on the graph of Part A. We begin by pushing the Y= button and typing the equation in the first three rows.

To see the graphs you will need to adjust the window. Push WINDOW, change the settings, and push GRAPH.

Next, to find the intersection push 2nd and TRACE. From this menu, choose intersect. The calculator will prompt you to choose the first and second curve and to provide the calculator with a best guess of where the intersection might be. You will need to repeat the process twice to find both intersection points.

According to the graph in Part A, if the value of x is between the two bounds we just found, then the cross section area is between 35 and 42 square inches. 6.6755812≤ x≤ 7.3127328 To answer the question we need to find a relationship between the value of x and the thickness of the drain pipe.