document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=158560>Question 158560</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/0470279923/algebrahomeworkh></A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> Mixture Problem.  A store sells almonds for $6 per pound, cashews for $5 per pound, and peanuts for $2 per pound.  One week the manager decides to prepare 100 16-ounce packages of nuts by mixing the peanuts, almonds, and cashews.  Each package will be sold for $4.  The mixture is to produce the same revenue as selling the nuts separately.  Prepare a table that shows some of the possible ways the manager can prepare the mixture.   </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #117073 by </B><A HREF=/tutors/aboutme.mpl?userid=ptaylor><B>ptaylor(2198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ptaylor><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=ptaylor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=117073\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> SEVERAL DAYS AGO, I STARTED WORKING ON THIS PROBLEM AND WAS CALLED AWAY.  I AM NOW BACK AND SEE THAT A SOLUTION HAS NOT BEEN PROVIDED. YOU PROBABLY HAVE ALREADY FOUND A SOLUTION BUT I THINK IT'S A GOOD PROBLEM, SO I'LL GIVE YOU MY TWO BITS:\r<BR>\n" );
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document.write( "This is yet another example of more unknowns than equations.  Many times, multiple answers are possible.<BR>\n" );
document.write( "First, lets deal in ounces:<BR>\n" );
document.write( "Let x=amount of almonds @ $6 per lb or $3/8 per ounce<BR>\n" );
document.write( "Let y=amount of cashews @ $5 per lb or $5/16 per ounce<BR>\n" );
document.write( "Let z=amount of peanuts @ $2 per lb or $1/8 per ounce \r<BR>\n" );
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document.write( "Now we are told that:<BR>\n" );
document.write( "x+y+z=16------------------------eq1<BR>\n" );
document.write( "and ($ are understood)<BR>\n" );
document.write( "(3/8)x+(5/16)y+(1/8)z=4 multiply each term by 16<BR>\n" );
document.write( "6x+5y+2z=64-----------------------------------------eq2\r<BR>\n" );
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document.write( "Multiply eq1 by 6 and subtract eq2 from it and we get:\r<BR>\n" );
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document.write( "y+4z=32  solve for y<BR>\n" );
document.write( "y=32-4z-------------------------------eq1a\r<BR>\n" );
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document.write( "Now we know a couple of things about this problem:<BR>\n" );
document.write( "(1) We cannot have negative values for x, y or z<BR>\n" );
document.write( "(2) We cannot have non-zero values for x, y, or z (I'm assuming that there has to be some of each in every packet)\r<BR>\n" );
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document.write( "Given the above, we see that in eq1a, 32-4z has to be greater than zero, other wise we will have either negative or a zero value for y, so:<BR>\n" );
document.write( "32-4z>0  subtract 32 from each side:<BR>\n" );
document.write( "-4z>-32 divide each side by -4 (note: the inequality sign changes)<BR>\n" );
document.write( "z<8  in order for y to be positive and non-zero<BR>\n" );
document.write( "Now lets plug y=32-4z from eq1a into eq1<BR>\n" );
document.write( "x+32-4z+z=16 or<BR>\n" );
document.write( "x+32-3z=16 subtract 32 from and add 3z to each side<BR>\n" );
document.write( "x=3z-16  now in order for x to be positive and non-zero, 3z-16 must be greater than zero:\r<BR>\n" );
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document.write( "3z-16>0  add 16 to each side<BR>\n" );
document.write( "3z>16  divide each side by 3<BR>\n" );
document.write( "z>5 1/3 in order for x to be positive and non-zero\r<BR>\n" );
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document.write( "WE HAVE NOW PLACED CONSTRAINTS ON Z:<BR>\n" );
document.write( "(5 1/3)< z <(8). In order for x and y to be positive and non-zero<BR>\n" );
document.write( "Now we can start building the table:<BR>\n" );
document.write( "Choose ANY value (INFINITE NUMBER DEPENDING ON THE ACCURACY OF THE SCALES) for z within the above constraints, plug the value into eq1a and solve for y, then plug both values into eq1 and solve for x.  Lets try an example:\r<BR>\n" );
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document.write( "Let z be 7, then from eq1a, y=4, then from eq1 x+4+7=16<BR>\n" );
document.write( "x+16-11; x=5\r<BR>\n" );
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document.write( "CK<BR>\n" );
document.write( "5+4+7=16<BR>\n" );
document.write( "6*5+5*4+7*2=64<BR>\n" );
document.write( "30+20+14=64<BR>\n" );
document.write( "64=64<BR>\n" );
document.write( "Niche problem!!!\r<BR>\n" );
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document.write( "Hope this helps---ptaylor\r<BR>\n" );
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document.write( " </SPAN>\n" );
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