SECTION 2.6 Mathematical Models: Building Functions 127 (c) If x 1, = the distance d is d 1 1 1 1 1 4 2 ( ) = − + = (d) If x 2 2 , = the distance d is d 2 2 2 2 2 2 1 1 4 1 2 1 3 2 4 2 ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ = ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ − ⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟ ⎟ + = − + = (e) Figure 68 shows the graph of d x x x 1 4 2 ( ) = − + using Desmos. By clicking on the two shadow points at the local minima, we find that when x 0.71 ≈ and 0.71 − , the value of d is smallest. The local minimum value is d 0.87 ≈ rounded to two decimal places. Since 0.71 1 0.50, 2 ( ) ± − ≈ − the points 0.71, 0.50 ( ) − − and 0.71, 0.50 ( ) − on the graph of y x 1 2 = − are closest to the origin. Figure 67 = − y x 1 2 x y 2 21 (0, 0) P5 (x, y) 21 1 2 1 d Area of a Rectangle A rectangle has one corner in quadrant I on the graph of y x 25 ,2 = − another at the origin, a third on the positive y-axis, and the fourth on the positive x-axis. See Figure 69. (a) Express the area A of the rectangle as a function of x. (b) What is the domain of A? (c) Graph A A x . ( ) = (d) For what value of x is the area A largest? EXAMPLE 2 Figure 68 ( ) = − + d x x x 1 4 2 Now Work PROBLEM 1 Solution (a) The area A of the rectangle is A xy, = where y x 25 .2 = − Substituting this expression for y, we obtain A x x x x x 25 25 . 2 3 ( ) ( ) = − = − (b) Since x y , ( ) is in quadrant I, we have x 0. > Also, y x 25 0, 2 = − > which implies that x 25, 2 < so x 5 5. − < < Combining these restrictions, the domain of A is x x |0 5 , { } < < or 0, 5 ( ) using interval notation. (c) See Figure 70 for the graph of A A x( ) = on a TI-84 Plus CE. (d) Using MAXIMUM, we find that the maximum area is 48.11 square units at x 2.89 = units, each rounded to two decimal places. See Figure 71. Figure 69 x y 10 20 30 21 (x, y) y 5 25 2 x2 4 5 1 2 3 (0, 0) Figure 70 ( ) = − A x x x 25 3 5 0 0 50 Figure 71 5 0 0 50 Now Work PROBLEM 7
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