254 CHAPTER 4 Polynomial and Rational Functions Figure 54 ( ) ( ) ( )( ) = + − + − Y x x x x 2 2 5 5 2 1 2 2 215 25 5 10 The numerator of a rational function ( ) ( ) ( ) = R x p x q x in lowest terms determines the x-intercepts of its graph. The graph shown in Figure 53 has x-intercepts −2 (even multiplicity; graph touches the x-axis) and 5 (odd multiplicity; graph crosses the x-axis). So one possibility for the numerator is ( ) ( ) ( ) = + − p x x x 2 5 . 2 The denominator of a rational function in lowest terms determines the vertical asymptotes of its graph. The vertical asymptotes of the graph are = − x 5 and = x 2. Since ( ) R x approaches ∞ to the left of = − x 5 and ( ) R x approaches −∞ to the right of = − x 5, we know that ( ) +x 5 is a factor of odd multiplicity in ( ) q x . Also, ( ) R x approaches −∞ on both sides of = x 2, so ( ) −x 2 is a factor of even multiplicity in ( ) q x . A possibility for the denominator is ( ) ( )( ) = + − q x x x 5 2 . 2 So far we have ( ) ( ) ( ) ( )( ) = + − + − R x x x x x 2 5 5 2 . 2 2 The horizontal asymptote of the graph given in Figure 53 is = y 2, so we know that the degree of the numerator must equal the degree of the denominator, which it does, and that the quotient of leading coefficients must be 2 1 . This leads to ( ) ( ) ( ) ( )( ) = + − + − R x x x x x 2 2 5 5 2 2 2 Solution Check: Figure 54 shows the graph of R on a TI-84 Plus CE. Since Figure 54 looks similar to Figure 53, we have found a rational function R for the graph in Figure 53. Now Work PROBLEM 51 2 Solve Applied Problems Involving Rational Functions Finding the Least Cost of a Can Reynolds Metal Company manufactures aluminum cans in the shape of a cylinder with a capacity of 500 cubic centimeters ( ) 1 2 liter . The top and bottom of the can are made of a special aluminum alloy that costs 0.05¢ per square centimeter. The sides of the can are made of material that costs 0.02¢ per square centimeter. (a) Express the cost of material for the can as a function of the radius r of the can. (b) Use a graphing utility to graph the function ( ) = C C r . (c) What value of r will result in the least cost? (d) What is this least cost? EXAMPLE 7 Figure 55 r r Top Bottom Area 5 pr 2 Area 5 pr 2 h h Lateral Surface Area 5 2prh Solution (a) Figure 55 illustrates the components of a can in the shape of a right circular cylinder. Notice that the material required to produce a cylindrical can of height h and radius r consists of a rectangle of area πrh 2 and two circles, each of area πr .2 The total cost C (in cents) of manufacturing the can is π π π π = + = ⋅ ⋅ + ⋅ = + C r rh r rh Cost of the top and bottom Cost of the side 2 0.05 2 0.02 0.10 0.04 2 2 There is an additional restriction that the height h and radius r must be chosen so that the volume V of the can is 500 cubic centimeters. Since π = V r h, 2 we have π π = = r h h r 500 so 500 2 2 Total area of top and bottom Cost/unit area Total area of side Cost/unit area
RkJQdWJsaXNoZXIy NjM5ODQ=