142 CHAPTER 1 Equations and Inequalities Geometry Problems To solve these applications, we continue to use a six-step problem-solving strategy. 1.5 Applications and Modeling with Quadratic Equations ■ Geometry Problems ■ The Pythagorean Theorem ■ Height of a Projected Object ■ Modeling with Quadratic Equations 5 5 5 5 5 5 5 5 3x – 10 x – 10 x 3x Figure 5 5 x – 10 3x – 10 Figure 6 EXAMPLE 1 Solving a Problem Involving Volume A piece of machinery produces rectangular sheets of metal such that the length is three times the width. Equal-sized squares measuring 5 in. on a side can be cut from the corners so that the resulting piece of metal can be shaped into an open box by folding up the flaps. If specifications call for the volume of the box to be 1435 in.3, find the dimensions of the original piece of metal. SOLUTION Step 1 Read the problem. We must find the dimensions of the original piece of metal. Step 2 Assign a variable. We know that the length is three times the width. Let x = the width (in inches) and thus, 3x = the length. The box is formed by cutting 5 + 5 = 10 in. from both the length and the width. See Figure 5. The width of the bottom of the box is x - 10, the length of the bottom of the box is 3x - 10, and the height is 5 in. (the length of the side of each cut-out square). See Figure 6. Step 3 Write an equation. The formula for the volume of a box is V = lwh. Volume = length * width * height 1435 = 13x - 1021x - 102152 (Note that the dimensions of the box must be positive numbers, so 3x - 10 and x - 10 must be greater than 0, which implies x 710 3 and x 710. These are both satisfied when x 710.) Step 4 Solve the equation from Step 3. 1435 = 15x2 - 200x + 500 Multiply. 0 = 15x2 - 200x - 935 Subtract 1435 from each side. 0 = 3x2 - 40x - 187 Divide each side by 5. 0 = 13x + 1121x - 172 Factor. 3x + 11 = 0 or x - 17 = 0 Zero-factor property x = - 11 3 or x = 17 Solve each equation. Step 5 State the answer. Only 17 satisfies the restriction x 710. Thus, the dimensions of the original piece should be 17 in. by 31172 = 51 in. Step 6 Check. The length and width of the bottom of the box are 51 - 2152 = 41 in. Length and 17 - 2152 = 7 in. Width The height is 5 in. (the amount cut on each corner), so the volume is V = lwh = 41172152 = 1435 in.3, as required. S Now Try Exercise 27. The width cannot be negative.
RkJQdWJsaXNoZXIy NjM5ODQ=