document.write( "Question 1194100: A charity worker is going to sell wigs, candles and hat clips to raise funds for their organization. The cost a wig for $10 and sell it at the price of $30, the cost of a candle is $ 4 and they sell them for $ 12 and the hat clips cost $ 10 and sell them for $24. They have the maximum budget to buy the stuff is $ 4000. the area of a table is 60\" long and 30\" wide. The area occupied by the wigs is 10square inch, the candle occupied 6square inch and the hat clips 2 square inch. How should the charity organize their stuff so they can get the maximum profit?
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Algebra.Com's Answer #826226 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Here is the step-by-step solution using the standard simplex method:\r\n" );
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document.write( "Suppose they will buy w wigs, c candles, and h hats, for a maximum profit P.\r\n" );
document.write( "[I used \"hats\" instead of \"hat clips\" for convenience]\r\n" );
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document.write( "                              |wigs |candles| hats|\r\n" );
document.write( "------------------------------|-----|-------|-----|------------\r\n" );
document.write( "How many bought and sold      |  w  |   c   |  h  |\r\n" );
document.write( "Cost each                     | 10  |   4   | 10  |\r\n" );
document.write( "Cost for all bought & sold.   |$10w | $4c   |$10h | ≤ $4000    <--cost \r\n" );
document.write( "Selling price for each        |$30  |$12    |$24  | \r\n" );
document.write( "Total money taken in          |$30w |$12c   |$24h | \r\n" );
document.write( "Profit for all bought & sold  |$20w | $8c   |$14h | = P        <--profit\r\n" );
document.write( "Sq. in. required              | 10w |  6c   |  2h | ≤ 1800     <--sq. in. \r\n" );
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document.write( "So the problem is:\r\n" );
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document.write( "Maximize P = 20w+8c+14h  <--objective function\r\n" );
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document.write( "subject to the constraints:\r\n" );
document.write( " 10w + 4c + 10h ≤ 4000\r\n" );
document.write( " 10w + 6c +  2h ≤ 1800\r\n" );
document.write( "w ≥ 0, c ≥ 0, h ≥ 0\r\n" );
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document.write( "Rewrite the objective function as\r\n" );
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document.write( "-20w - 8c - 14h + P = 0\r\n" );
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document.write( "Introduce slack variables s1 and s2 to change the inequalities \r\n" );
document.write( "into equations, and put the rewritten objective function at \r\n" );
document.write( "the bottom:\r\n" );
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document.write( "Next, we put the above system in a special 3x7 partitioned \r\n" );
document.write( "matrix called a \"tableau\", with each variable written \r\n" );
document.write( "above its column of coefficients.  The constants in the\r\n" );
document.write( "right-most column are headed k:\r\n" );
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document.write( "|  w  c   h |s1 s2 | P |  k   |\r\n" );
document.write( "-------------------------------  \r\n" );
document.write( "| 10  4  10 | 1  0 | 0 | 4000 |\r\n" );
document.write( "| 10  6   2 | 0  1 | 0 | 1800 |\r\n" );
document.write( "-------------------------------\r\n" );
document.write( "|-20 -8 -14 | 0  0 | 1 |    0 |\r\n" );
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document.write( "We find the most negative number ('indicator') on the \r\n" );
document.write( "bottom row, which is -20.  Its column, the 1st column,\r\n" );
document.write( "is the \"pivot\" column.\r\n" );
document.write( "We divide each POSITIVE number in the k-column by the\r\n" );
document.write( "POSITIVE number in the pivot column which is in the \r\n" );
document.write( "same row.  That is:\r\n" );
document.write( "4000/10=400, 1800/10=180\r\n" );
document.write( "Since 180 is smaller, the pivot row is the 2nd row.\r\n" );
document.write( "The pivot element is the 10 in the 2nd row. \r\n" );
document.write( "We use row operations to make the pivot element become\r\n" );
document.write( "1 and all the other elements in the pivot column become 0.\r\n" );
document.write( "Here are three row operations we perform on the above\r\n" );
document.write( "tableau matrix:\r\n" );
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document.write( "(1/10)R2 -> R2\r\n" );
document.write( "-10R2+R1 -> R1\r\n" );
document.write( "20R2+R3 -> R3\r\n" );
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document.write( "|  w  c   h |s1 s2 | P |  k   |\r\n" );
document.write( "-------------------------------  \r\n" );
document.write( "|  0 -2   8 | 1 -1 | 0 | 2200 |\r\n" );
document.write( "|  1 .6  .2 | 0 .1 | 0 |  180 |\r\n" );
document.write( "-------------------------------\r\n" );
document.write( "|  0  4 -10 | 0  2 | 1 | 3600 |\r\n" );
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document.write( "We still have a negative number on the bottom row, -10.\r\n" );
document.write( "Now we find the most negative number ('indicator') on the \r\n" );
document.write( "bottom row. It happens to be the ONLY negative number on\r\n" );
document.write( "the bottom row. which is -10.  Its column, the 3rd column,\r\n" );
document.write( "is the \"pivot\" column.\r\n" );
document.write( "We divide each POSITIVE number in the k-column by each\r\n" );
document.write( "POSITIVE number in the pivot column which is in the \r\n" );
document.write( "same row:\r\n" );
document.write( "2200/8=400, 180/.2=1800\r\n" );
document.write( "Since 400 is smaller, the pivot row is the 1st row.\r\n" );
document.write( "The pivot element is the 8 in the 1st row. \r\n" );
document.write( "We use row operations to make the pivot element become\r\n" );
document.write( "1 and all the other elements in the pivot column become 0.\r\n" );
document.write( "Here are the row operations we perform on the above:\r\n" );
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document.write( "(1/8)R1 -> R1\r\n" );
document.write( "-.2R1+R2 -> R2\r\n" );
document.write( "10R1+R3 -> R3\r\n" );
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document.write( "|  w    c   h |  s1    s2   | P |   k  |\r\n" );
document.write( "----------------------------------------  \r\n" );
document.write( "|  0 -.25   1 | .125 -.125  | 0 |  275 |\r\n" );
document.write( "|  1  .65   0 |-.025  .125  | 0 |  125 |\r\n" );
document.write( "----------------------------------------\r\n" );
document.write( "|  0  1.5   0 | 1.25   .75  | 1 | 6350 |\r\n" );
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document.write( "This tableau is finished because there are\r\n" );
document.write( "no more negative numbers (indicators) on \r\n" );
document.write( "the bottom row.\r\n" );
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document.write( "We transform the tableau matrix back into\r\n" );
document.write( "a system of equations:\r\n" );
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document.write( "That simplifies to:\r\n" );
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document.write( "Solve the bottom equation for P\r\n" );
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document.write( "\"1.5c%2B1.25s%5B1%5D%2B0.75s%5B2%5D%2BP=6350\"\r\n" );
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document.write( "\"P=6350-1.5c-1.25s%5B1%5D-0.75s%5B2%5D\"\r\n" );
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document.write( "Since c, s1 and s2 cannot be negative, we see that\r\n" );
document.write( "the largest value P can take on is P = 6350 when \r\n" );
document.write( "c = s1 = s2 = 0.  So they will not buy any candles!!\r\n" );
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document.write( "So we substitute in the other two equations to find\r\n" );
document.write( "the number of wigs and hats:\r\n" );
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document.write( "h = 275, w = 125\r\n" );
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document.write( "So the most lucrative plan is to buy \r\n" );
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document.write( "125 wigs, 0 candles, and 275 hats. and the maximum profit P = $6350 \r\n" );
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