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Stitz-Zeager_College_Algebra_e-book

# In part 2 we set f x g x and solved to nd x 3 for

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Unformatted text preview: the graph y = P (x) is above the x-axis, meaning y = P (x) > 0 there. This means that for these values of x, a proﬁt is being made. Since x represents the weekly sales of PortaBoy Game Systems, we round the zeros to positive integers and have that as long as 1, but no more than 112 game systems are sold weekly, the retailer will make a proﬁt. 146 Linear and Quadratic Functions 4. From the graph, we see the maximum value of P occurs at the vertex, which is approximately (56.67, 4666.67). As above, x represents the weekly sales of PortaBoy systems, so we can’t sell 56.67 game systems. Comparing P (56) = 4666 and P (57) = 4666.5, we conclude we will make a maximum proﬁt of \$4666.50 if we sell 57 game systems. 5. In the previous part, we found we need to sell 57 PortaBoys per week to maximize proﬁt. To ﬁnd the price per PortaBoy, we substitute x = 57 into the price-demand function to get p(57) = −1.5(57) + 250 = 164.5. The price should be set at \$164.50. We conclude this section with a more complicated absolute va...
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