document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1170424>Question 1170424</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝟏 <BR>\n" );
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document.write( "(a)  Teddy J is a manufacturer of dish washing liquid .  If his monthly demand function  for 750ml  size is   q = 4000 − 250p  and  his total cost function is  C(q) = 500 + 0.2q.    <BR>\n" );
document.write( "(i)	Derive an expression,   R(q) for  Teddy J′s total revenue curve. <BR>\n" );
document.write( "(ii)	Derive an expression,    Π(q) for Teddy J′s profit function. <BR>\n" );
document.write( "(iii)	Determine whether  Teddy J′s profit is increasing or decreasing when  he produces  5 hundred, 750ml bottles of dish washing liquid. <BR>\n" );
document.write( "(iv)	How many 750ml bottles of dish washing liquid should Teddy J produce  per  month if he wishes to maximize his profits. <BR>\n" );
document.write( "(b)    A firm has an average cost function <BR>\n" );
document.write( "	125	q2<BR>\n" );
document.write( "A(q) = + − 4.     q	16<BR>\n" );
document.write( "where q is the firm′s  output. <BR>\n" );
document.write( "(i)	Determine the level of output for average costs are minimum. <BR>\n" );
document.write( "(ii)	Hence determine the range of values for which average costs are decreasing. <BR>\n" );
document.write( "(iii)	What part of the decreasing range is practically feasible? <BR>\n" );
document.write( "(iv)	Write an equation for the total cost function. <BR>\n" );
document.write( "(v)	Hence calculate the level of output for which total costs are minimum.   <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 #851189 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=851189\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's break down each part of the problem.\r<BR>\n" );
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document.write( "**(a) Teddy J's Dish Washing Liquid**\r<BR>\n" );
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document.write( "Given:<BR>\n" );
document.write( "* Demand function: $q = 4000 - 250p$<BR>\n" );
document.write( "* Total cost function: $C(q) = 500 + 0.2q$\r<BR>\n" );
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document.write( "**(i) Total Revenue (R(q))**\r<BR>\n" );
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document.write( "* First, solve the demand function for $p$:<BR>\n" );
document.write( "    * $250p = 4000 - q$<BR>\n" );
document.write( "    * $p = \frac{4000 - q}{250} = 16 - \frac{q}{250}$<BR>\n" );
document.write( "* Total revenue is given by $R(q) = pq$:<BR>\n" );
document.write( "    * $R(q) = \left(16 - \frac{q}{250}\right)q$<BR>\n" );
document.write( "    * $R(q) = 16q - \frac{q^2}{250}$\r<BR>\n" );
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document.write( "**(ii) Profit Function (Π(q))**\r<BR>\n" );
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document.write( "* Profit is total revenue minus total cost:<BR>\n" );
document.write( "    * $\Pi(q) = R(q) - C(q)$<BR>\n" );
document.write( "    * $\Pi(q) = \left(16q - \frac{q^2}{250}\right) - (500 + 0.2q)$<BR>\n" );
document.write( "    * $\Pi(q) = 16q - \frac{q^2}{250} - 500 - 0.2q$<BR>\n" );
document.write( "    * $\Pi(q) = -\frac{q^2}{250} + 15.8q - 500$\r<BR>\n" );
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document.write( "**(iii) Profit at q = 500**\r<BR>\n" );
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document.write( "* To determine if profit is increasing or decreasing, we need to find the derivative of the profit function and evaluate it at $q = 500$:<BR>\n" );
document.write( "    * $\Pi'(q) = -\frac{2q}{250} + 15.8 = -\frac{q}{125} + 15.8$<BR>\n" );
document.write( "    * $\Pi'(500) = -\frac{500}{125} + 15.8 = -4 + 15.8 = 11.8$<BR>\n" );
document.write( "* Since $\Pi'(500) > 0$, the profit is increasing when Teddy J produces 500 bottles.\r<BR>\n" );
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document.write( "**(iv) Profit Maximization**\r<BR>\n" );
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document.write( "* To maximize profit, we set the derivative of the profit function to zero:<BR>\n" );
document.write( "    * $\Pi'(q) = -\frac{q}{125} + 15.8 = 0$<BR>\n" );
document.write( "    * $\frac{q}{125} = 15.8$<BR>\n" );
document.write( "    * $q = 15.8 \cdot 125 = 1975$<BR>\n" );
document.write( "* Teddy J should produce 1975 bottles to maximize profits.\r<BR>\n" );
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document.write( "**(b) Firm's Average Cost Function**\r<BR>\n" );
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document.write( "Given:<BR>\n" );
document.write( "* Average cost function: $A(q) = \frac{125}{q} + \frac{q}{16} - 4$\r<BR>\n" );
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document.write( "**(i) Output for Minimum Average Costs**\r<BR>\n" );
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document.write( "* To find the minimum average cost, we take the derivative of $A(q)$ and set it to zero:<BR>\n" );
document.write( "    * $A'(q) = -\frac{125}{q^2} + \frac{1}{16} = 0$<BR>\n" );
document.write( "    * $\frac{125}{q^2} = \frac{1}{16}$<BR>\n" );
document.write( "    * $q^2 = 125 \cdot 16 = 2000$<BR>\n" );
document.write( "    * $q = \sqrt{2000} = 20\sqrt{5} \approx 44.72$<BR>\n" );
document.write( "* The level of output for minimum average costs is approximately 44.72.\r<BR>\n" );
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document.write( "**(ii) Range for Decreasing Average Costs**\r<BR>\n" );
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document.write( "* Average costs are decreasing when $A'(q) < 0$:<BR>\n" );
document.write( "    * $-\frac{125}{q^2} + \frac{1}{16} < 0$<BR>\n" );
document.write( "    * $\frac{1}{16} < \frac{125}{q^2}$<BR>\n" );
document.write( "    * $q^2 < 125 \cdot 16 = 2000$<BR>\n" );
document.write( "    * $q < \sqrt{2000} = 20\sqrt{5} \approx 44.72$<BR>\n" );
document.write( "* Average costs are decreasing when $0 < q < 20\sqrt{5}$.\r<BR>\n" );
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document.write( "**(iii) Practically Feasible Decreasing Range**\r<BR>\n" );
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document.write( "* Output must be a positive value.<BR>\n" );
document.write( "* The practically feasible range for decreasing average costs is $0 < q < 20\sqrt{5}$.\r<BR>\n" );
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document.write( "**(iv) Total Cost Function (C(q))**\r<BR>\n" );
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document.write( "* Total cost is average cost multiplied by output:<BR>\n" );
document.write( "    * $C(q) = q \cdot A(q)$<BR>\n" );
document.write( "    * $C(q) = q \left(\frac{125}{q} + \frac{q}{16} - 4\right)$<BR>\n" );
document.write( "    * $C(q) = 125 + \frac{q^2}{16} - 4q$\r<BR>\n" );
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document.write( "**(v) Output for Minimum Total Costs**\r<BR>\n" );
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document.write( "* To find the minimum total cost, we take the derivative of $C(q)$ and set it to zero:<BR>\n" );
document.write( "    * $C'(q) = \frac{2q}{16} - 4 = \frac{q}{8} - 4 = 0$<BR>\n" );
document.write( "    * $\frac{q}{8} = 4$<BR>\n" );
document.write( "    * $q = 32$<BR>\n" );
document.write( "* The level of output for minimum total costs is 32.<BR>\n" );
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