document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1180889>Question 1180889</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A company runs food device concessions for sporting events throughout the country their marketing research department chose a particular football stadium to test market a new jumbo hot dog. It was found that the demand for the new hot dog is given approximately by : p=4-ln(x),5 is less than equal to x and x is less than equal to 500.<BR>\n" );
document.write( "Where x is the number of hot dogs(in thousands) that can be sold during one game at a price of P dollars. If the company pays 1 dollar for each hot dog, how should hot dogs be priced to maximize the profit per game?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850091 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=850091\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to determine the optimal price to maximize profit:\r<BR>\n" );
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document.write( "**1. Define the Profit Function:**\r<BR>\n" );
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document.write( "* **Revenue:** Revenue is price (p) times quantity (x):  R(x) = p * x = (4 - ln(x)) * x<BR>\n" );
document.write( "* **Cost:** The cost is $1 per hot dog times the quantity: C(x) = 1 * x = x<BR>\n" );
document.write( "* **Profit:** Profit is revenue minus cost: P(x) = R(x) - C(x) = (4 - ln(x))x - x = 4x - x*ln(x) - x = 3x - x*ln(x)\r<BR>\n" );
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document.write( "**2. Find the Derivative of the Profit Function:**\r<BR>\n" );
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document.write( "To find the maximum profit, we need to find the critical points of the profit function by taking its derivative and setting it to zero:\r<BR>\n" );
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document.write( "P'(x) = d(3x - x*ln(x))/dx = 3 - (ln(x) + x*(1/x)) = 3 - ln(x) - 1 = 2 - ln(x)\r<BR>\n" );
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document.write( "**3. Set the Derivative Equal to Zero and Solve for x:**\r<BR>\n" );
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document.write( "P'(x) = 0<BR>\n" );
document.write( "2 - ln(x) = 0<BR>\n" );
document.write( "ln(x) = 2<BR>\n" );
document.write( "x = e² ≈ 7.389\r<BR>\n" );
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document.write( "Since x is in thousands, this means approximately 7389 hot dogs.\r<BR>\n" );
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document.write( "**4. Verify that this is a Maximum (Second Derivative Test):**\r<BR>\n" );
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document.write( "Find the second derivative of the profit function:\r<BR>\n" );
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document.write( "P''(x) = d(2 - ln(x))/dx = -1/x\r<BR>\n" );
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document.write( "Since x is always positive in our domain (5 ≤ x ≤ 500), P''(x) is always negative. A negative second derivative indicates a maximum.\r<BR>\n" );
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document.write( "**5. Find the Optimal Price:**\r<BR>\n" );
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document.write( "Substitute the value of x back into the demand equation to find the optimal price:\r<BR>\n" );
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document.write( "p = 4 - ln(e²)<BR>\n" );
document.write( "p = 4 - 2<BR>\n" );
document.write( "p = 2\r<BR>\n" );
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document.write( "**6. Check the endpoints:**\r<BR>\n" );
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document.write( "Since the domain of x is restricted (5 ≤ x ≤ 500), we also need to check the profit at the endpoints:\r<BR>\n" );
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document.write( "* **x = 5:** P(5) = 3(5) - 5ln(5) ≈ 15 - 8.047 ≈ 6.953<BR>\n" );
document.write( "* **x = 500:** P(500) = 3(500) - 500ln(500) ≈ 1500 - 3224 ≈ -1724\r<BR>\n" );
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document.write( "The profit at x = e² is: P(e²) = 3e² - e²ln(e²) = 3e² - 2e² = e² ≈ 7.389\r<BR>\n" );
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document.write( "**Conclusion:**\r<BR>\n" );
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document.write( "The company should price the jumbo hot dogs at $2 to maximize profit. This will result in selling approximately 7389 hot dogs.<BR>\n" );
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