172 CHAPTER 3 Linear and Quadratic Functions Figure 26 (0, 0) 56 14 28 42 Price per calculator (dollars) 70 84 98 112 126 140 800,000 700,000 600,000 500,000 Revenue (dollars) 400,000 200,000 100,000 300,000 p R (50, 675 000) (70, 735 000) (90, 675 000) Figure 25 (70, 735 000) Price per calculator (dollars) 800,000 700,000 600,000 500,000 400,000 200,000 100,000 300,000 (0, 0) 14 28 42 56 70 84 98 112 126 140 p R Revenue (dollars) Now Work PROBLEM 3 (c) The function R is a quadratic function with a b 150, 21,000, = − = and c 0. = Because a 0, < the vertex is the highest point on the parabola. The revenue R is a maximum when the price p is p b a2 21,000 2 150 $70.00 ( ) = − = − − = ↑ =− = a b 150, 21,000 (d) The maximum revenue R is R 70 150 70 21,000 70 $735,000 2 ( ) = − ⋅ + ⋅ = (e) The number of calculators sold is given by the demand equation x p 21,000 150 . = − At a price of p $70, = x 21,000 150 70 10,500 = − ⋅ = calculators are sold. (f) To graph R, plot the intercepts 0, 0 ( ) and 140, 0 ( ) and the vertex 70, 735 000 . ( ) See Figure 25 for the graph. Add − p p 150 21,000 2 to both sides. Divide both sides by 150. Factor. Use the Zero-Product Property. (e) Graph R 675,000 = and R p p p 150 21,000 2 ( ) = − + on the same Cartesian plane. See Figure 25. We find where the graphs intersect by solving p p p p p p p p p p 675,000 150 21,000 150 21,000 675,000 0 140 4500 0 50 90 0 50 or 90 2 2 2 ( )( ) = − + − + = − + = − − = = = The graphs intersect at 50, 675 000 ( ) and 90, 675 000 . ( ) Based on the graph in Figure 26, Texas Instruments should charge between $50 and $90 to earn at least $675,000 in revenue.
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