SECTION 3.4 Building Quadratic Models from Verbal Descriptions and from Data 173 Figure 27 x x w w Maximizing the Area Enclosed by a Fence A farmer has 2000 yards of fence to enclose a rectangular field. What are the dimensions of the rectangle that encloses the most area? EXAMPLE 2 Solution Figure 27 illustrates the situation.The available fence represents the perimeter of the rectangle. If x is the length and w is the width, then x w 2 2 2000 + = (1) The area A of the rectangle is A xw = To express A in terms of a single variable, solve equation (1) for w and substitute the result in A xw. = Then A involves only the variable x. [You could also solve equation (1) for x and express A in terms of w alone. Try it!] x w w x w x x 2 2 2000 2 2000 2 2000 2 2 1000 + = = − = − = − Then the area A is A xw x x x x 1000 1000 2 ( ) = = − = − + Now, A is a quadratic function of x. A A x x x 1000 2 ( ) = = − + =− = = a b c 1, 1000, 0 Figure 28 shows the graph of A x x x 1000 . 2 ( ) = − + Because a 0, < the vertex is a maximum point on the graph of A. The maximum value occurs at x b a2 1000 2 1 500 ( ) = − = − − = The maximum value of A is A b a A 2 500 500 1000 500 250,000 500,000 250,000 2 ( ) ( ) − = = − + ⋅ = − + = The largest rectangle that can be enclosed by 2000 yards of fence has an area of 250,000 square yards. Its dimensions are 500 yards by 500 yards. *A cable suspended from two towers is in the shape of a catenary, but when a horizontal roadway is suspended from the cable, the cable takes the shape of a parabola. Now Work PROBLEM 7 Figure 28 A x x x 1000 2 ( ) =− + (500, 250 000) (0, 0) (1000, 0) A 500 1000 250,000 x The Golden Gate Bridge The Golden Gate Bridge, a suspension bridge, spans the entrance to San Francisco Bay. Its 746-foot-tall towers are 4200 feet apart. The bridge is suspended from two huge cables more than 3 feet in diameter; the 90-foot-wide roadway is 220 feet above the water.The cables are parabolic in shape* and touch the road surface at the center of the bridge. Find the height of the cable above the road at a distance of 1000 feet from the center. Solution EXAMPLE 3 Begin by choosing the coordinate axes so that the x-axis coincides with the road surface and the origin coincides with the center of the bridge. See Figure 29. As a result, the 746-foot towers will be vertical (height 746 220 526 feet − = above the road) and located 2100 feet from the center. Also, the cable has the shape of a parabola that is concave up.The parabola extends from the towers and has its vertex at 0, 0 . ( ) This choice of the axes results in the equation of the parabola having the form y ax a , 0. 2 = > Note that the points 2100, 526 ( ) − and 2100, 526 ( ) are on the graph. Figure 29 220' (0, 0) 746' 526' 2100' 2100' ? (2100, 526) (22100, 526) 1000' y x (continued)
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