document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1174395>Question 1174395</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Factory A has produced a product for the common users. The total variable cost per unit is<BR>\n" );
document.write( "RM 10. The fixed cost is RM 6 200 in a year and the yearly revenue (in RM) is derived by<BR>\n" );
document.write( "function R(x) = R(x) =  -  2x^2+260x+500, given that x is the number of productions.\r<BR>\n" );
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document.write( "(a) Find the production level for breakeven point of factory A.\r<BR>\n" );
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document.write( "(b) State the range of number of productions where the factory A should produce to have profit.\r<BR>\n" );
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document.write( "(c) State the range of number of productions when the factory A has a loss more than RM 500. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850691 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850691\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's break down this problem step-by-step.\r<BR>\n" );
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document.write( "**(a) Breakeven Point**\r<BR>\n" );
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document.write( "1.  **Cost Function:**<BR>\n" );
document.write( "    * Variable Cost (VC) = 10x<BR>\n" );
document.write( "    * Fixed Cost (FC) = 6200<BR>\n" );
document.write( "    * Total Cost (C(x)) = VC + FC = 10x + 6200\r<BR>\n" );
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document.write( "2.  **Revenue Function:**<BR>\n" );
document.write( "    * R(x) = -2x² + 260x + 500\r<BR>\n" );
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document.write( "3.  **Breakeven Point:**<BR>\n" );
document.write( "    * Breakeven occurs when Revenue (R(x)) = Cost (C(x)).<BR>\n" );
document.write( "    * -2x² + 260x + 500 = 10x + 6200<BR>\n" );
document.write( "    * -2x² + 250x - 5700 = 0<BR>\n" );
document.write( "    * Divide by -2: x² - 125x + 2850 = 0\r<BR>\n" );
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document.write( "4.  **Solve the Quadratic Equation:**<BR>\n" );
document.write( "    * We can use the quadratic formula: x = [-b ± √(b² - 4ac)] / 2a<BR>\n" );
document.write( "    * In this case, a = 1, b = -125, c = 2850.<BR>\n" );
document.write( "    * x = [125 ± √((-125)² - 4 * 1 * 2850)] / 2<BR>\n" );
document.write( "    * x = [125 ± √(15625 - 11400)] / 2<BR>\n" );
document.write( "    * x = [125 ± √4225] / 2<BR>\n" );
document.write( "    * x = [125 ± 65] / 2\r<BR>\n" );
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document.write( "    * x1 = (125 + 65) / 2 = 190 / 2 = 95<BR>\n" );
document.write( "    * x2 = (125 - 65) / 2 = 60 / 2 = 30\r<BR>\n" );
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document.write( "    * **Answer (a):** The breakeven points are at production levels of 30 units and 95 units.\r<BR>\n" );
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document.write( "**(b) Range of Production for Profit**\r<BR>\n" );
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document.write( "1.  **Profit Function:**<BR>\n" );
document.write( "    * Profit (P(x)) = R(x) - C(x)<BR>\n" );
document.write( "    * P(x) = (-2x² + 260x + 500) - (10x + 6200)<BR>\n" );
document.write( "    * P(x) = -2x² + 250x - 5700\r<BR>\n" );
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document.write( "2.  **Profit Occurs When P(x) > 0:**<BR>\n" );
document.write( "    * -2x² + 250x - 5700 > 0<BR>\n" );
document.write( "    * x² - 125x + 2850 < 0 (divide by -2 and reverse the inequality)\r<BR>\n" );
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document.write( "3.  **Use the Breakeven Points:**<BR>\n" );
document.write( "    * We already found the breakeven points at x = 30 and x = 95.<BR>\n" );
document.write( "    * Since the parabola opens downward (-2x²), the profit is positive between the breakeven points.\r<BR>\n" );
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document.write( "    * **Answer (b):** The factory should produce between 30 and 95 units to have a profit (30 < x < 95).\r<BR>\n" );
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document.write( "**(c) Range of Production for Loss > RM 500**\r<BR>\n" );
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document.write( "1.  **Loss Function:**<BR>\n" );
document.write( "    * Loss occurs when P(x) < 0.<BR>\n" );
document.write( "    * We want to find when P(x) < -500.\r<BR>\n" );
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document.write( "2.  **Set Up Inequality:**<BR>\n" );
document.write( "    * -2x² + 250x - 5700 < -500<BR>\n" );
document.write( "    * -2x² + 250x - 5200 < 0<BR>\n" );
document.write( "    * x² - 125x + 2600 > 0\r<BR>\n" );
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document.write( "3.  **Solve the Quadratic Equation:**<BR>\n" );
document.write( "    * x² - 125x + 2600 = 0<BR>\n" );
document.write( "    * Use the quadratic formula:<BR>\n" );
document.write( "        * x = [125 ± √((-125)² - 4 * 1 * 2600)] / 2<BR>\n" );
document.write( "        * x = [125 ± √(15625 - 10400)] / 2<BR>\n" );
document.write( "        * x = [125 ± √5225] / 2<BR>\n" );
document.write( "        * x = [125 ± 72.28] / 2\r<BR>\n" );
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document.write( "    * x1 = (125 + 72.28) / 2 ≈ 98.64<BR>\n" );
document.write( "    * x2 = (125 - 72.28) / 2 ≈ 26.36\r<BR>\n" );
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document.write( "4.  **Determine the Range:**<BR>\n" );
document.write( "    * Since the parabola opens upward, the inequality is satisfied when x < 26.36 or x > 98.64.\r<BR>\n" );
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document.write( "    * **Answer (c):** The factory has a loss greater than RM 500 when the number of productions is less than 26.36 units or greater than 98.64 units (x < 26.36 or x > 98.64).<BR>\n" );
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