document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=935446>Question 935446</A>: <I><SPAN STYLE=\"word-break: break-all;\"> callenge:hard<BR>\n" );
document.write( "A shoe maker sells a new model of athletic shoe for $90. On average, the store<BR>\n" );
document.write( "sells 45 pairs of these shoes each week. Jason, the sale manager, estimates that<BR>\n" );
document.write( "he will sell 4 additional pairs of shoes each week for every $5 reduction in the<BR>\n" );
document.write( "price .Find the following<BR>\n" );
document.write( "1- A quadratic function modeling the store’s average weekly revenue from the<BR>\n" );
document.write( "sale of these shoes as related to reduction in price.<BR>\n" );
document.write( "2- The range of price which would allow he store to maintain or increase its<BR>\n" );
document.write( "current weekly revenue from these shoes.<BR>\n" );
document.write( "3- The maximum amount by which the store can increase its revenue from these<BR>\n" );
document.write( "shoes each week . Round of to the nearest whole dollar.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #568679 by </B><A HREF=/tutors/aboutme.mpl?userid=DrBeeee><B>DrBeeee(684)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=DrBeeee><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=568679\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let n = the integer number of $5 reduction in price of the athletic shoes.<BR>\n" );
document.write( "Let I = the total weekly income (revenue).<BR>\n" );
document.write( "We have an income of the number of shoes sold times the price of the shoes <BR>\n" );
document.write( "(1) I = (45 + 4n)*(90 - 5n) <BR>\n" );
document.write( "Simplifying (1) we get<BR>\n" );
document.write( "(2) I = 45*90 - 45*5n + 4*90n -4*5n^2 or<BR>\n" );
document.write( "(3) I = 4050 - 225n + 360n - 20n^2 or<BR>\n" );
document.write( "(4) I = 4050 + 135n - 20n^2 or<BR>\n" );
document.write( "(5) I = - 20n^2 + 135n + 4050<BR>\n" );
document.write( "Equation (5) is the quadratic solution asked for in part 1.<BR>\n" );
document.write( "The answer to part 2 is that value of n at which the value of I goes below 4050. We need to set (5) equal to zero and find the two values of n that make (5) equal to 4050. We can see that I of (5) is equal to 4050 when do not change the price, i.e. n = 0. At n=0 (5) becomes<BR>\n" );
document.write( "(6) -20*0 + 135*0 + 4050 = 4050<BR>\n" );
document.write( "The quadratic of (5) increases as n increases, however when the first two term sum to a negative value we get less than 4050. This happens when <BR>\n" );
document.write( "(7) -20n^2 + 135n = 0 or<BR>\n" );
document.write( "(8) n*(-20n + 135) = 0 or<BR>\n" );
document.write( "(9) n = {0,6.75)<BR>\n" );
document.write( "Answer to part 2; The price range is (90, 90-5*6.75) or ($90,$56.25)<BR>\n" );
document.write( "Part 3 asks for the maximum value of I given by (5). This occurs when the derivative of (5) equals zero (if you're into calculus) or the apex of the parabola for algebra students. The answer is the same<BR>\n" );
document.write( "(10) Imax occurs at n = 135/(2*20) or<BR>\n" );
document.write( "(11) n = 3.375<BR>\n" );
document.write( "Put this value of n into (5) to get<BR>\n" );
document.write( "(12) Imax = -20n^2 + 135n + 4050 or<BR>\n" );
document.write( "(13) Imax = -20(3.375)^2 + 135*3.375 + 4050 or<BR>\n" );
document.write( "(14) Imax = -20n^2 + 135n + 4050 or<BR>\n" );
document.write( "(15) Imax = -227.8125 + 455.625 + 4050 or<BR>\n" );
document.write( "(16) Imax = 227.8125 + 4050 or<BR>\n" );
document.write( "(17) Imax = 4277.8125<BR>\n" );
document.write( "Answer to part 3; The maximum weekly revenue is $4278.\r<BR>\n" );
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