0. Code

>	restart; with(plots):

Warning, the name changecoords has been redefined

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MixturePlot  :=  proc( a, ar, b, br)      local     c,ct,cr,g,as,bs,cs, spacer, bo, co, pa, pb, pc1, pc2, tp, te, r, ch;      c := a + b;      ct := a*ar + b*br;  cr := ct/c;      g := 1.6;      as := sqrt( a/g);   bs := sqrt( b/g);  cs := sqrt( c/g);      spacer := (as+bs+cs)/10;             pa := polygonplot( [ [0,0],[as,0],[as,as*g],[0,as*g] ],                         color = gold, axes = none):      bo := as + spacer;             pb := polygonplot( [ [bo, 0],[bo+bs, 0],[bo+bs,bs*g],[bo,bs*g] ],                         color = green, axes = none):      co := bo + spacer*1.5 + bs;         r := a/(a+b);  ch := r*cs*g;      pc1 := polygonplot( [ [co,0],[co+cs,0],[co+cs,ch],[co,ch] ],                         color = gold, axes = none):      pc2 := polygonplot( [ [co,ch],[co+cs,ch],[co+cs,cs*g],[co,cs*g] ],                         color = green, axes = none):      tp := textplot( [ as + spacer/2, max(as,bs)*g/2, `+`]);      te := textplot( [ bo + bs + spacer/2, max(as,bs)*g/2, `=`]);      display( pa, pb, pc1, pc2, tp, te)  end proc:

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MixtureTableTest  :=  proc( a, ar, b, br )      # a = quanity of a, ar = a's rate, and so forth for b      local    rt, M ;      rt := (a*ar + b*br)/(a+b);      if( not(  (whattype(rt) = integer)  or                (whattype(rt) = float)    or                (whattype(rt) = fraction) )) then rt := `?` fi;      array( [ [``, `Quantity   *`,` Rate`,` =  Result`],               [ `A`,   a,   ar,                a*ar],               [ `B`,   b,   br,                b*br],               [ `A+B  `, a+b, rt, a*ar + b*br] ]);  end proc:

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MixtureTable1  :=  proc( a, ar, b, br, cr )      # a = quanity of a, ar = a's rate, and so forth for b      local    rt, M, equation;      rt := (a*ar + b*br)/(a+b);      if( not(  (whattype(rt) = integer)  or                (whattype(rt) = float)    or                (whattype(rt) = fraction) )) then rt := `?` fi;      M := array( [ [``, `Quantity   *`,` Rate`,` =  Result`],               [ `A`,   a,   ar,                a*ar],               [ `B`,   b,   br,                b*br],               [ `A+B  `, a+b, cr, (a+b)*cr] ]);      print(M);      print(` `):      equation  := M[2,4] + M[3,4]  = M[4,4];      print( equation   );      solve( equation, x); print(x = %);  end proc:

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MixtureTableEq  :=  proc( ar, br, cr, c)      #||a rate|b rate|c rate|c amount||          local    a, b, rt, M, equation;      rt := cr ; #(a*ar + b*br)/(a+b);      a := x;   b := c-a;                M := array( [ [``,  `Quantity   *`, ` Rate`,` =  Result`],                   [ `A`,     a,     ar,    a*ar        ],                 [ `B`,     b,   br,    (c-a)*br    ],                 [ `A+B  `, c,     cr,    c*cr        ] ]);      print(M);      print(` `):      equation  := M[2,4] + M[3,4]  = M[4,4];      print( equation );      solve( equation, x); print(` `); print(A = %, B = c-%); print(` `);        end proc:

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MixtureTableSoln  :=  proc( ar, br, cr, c)  #||a rate|b rate|c rate|c amount||           local    a, rt, M, equation;      rt := cr ; #(a*ar + b*br)/(a+b);      a := solve( x*ar + (c-x)*br = c*cr, x);                array( [ [``,  `Quantity   *`, ` Rate`,` =  Result`],                   [ `A`,     a,     ar,    a*ar        ],                 [ `B`,     c-a,   br,    (c-a)*br    ],                 [ `A+B  `, c,     cr,    c*cr        ] ]);    end proc:

1. General Idea

There are many variations of mixture problems such as mxitures of coffee, nuts, and coins. First, lets just get a rough idea of what is going on.

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Problem 1.1 :  You mix 5 pounds of nuts worth $2/lb and 3 pounds of nuts worth $4/lb.

>	MixturePlot( 5, 2, 3, 4);    #(a amount) | (a rate) | (b amount) | (b rate)

There are three items we can compute :
        1. How many pounds are in the mixture?
        2. How much does the entire mixture cost?
        3. How much is the mixture worth per pound?

>	`Total number of pounds` := 5 + 3; `Total cost` := 5*2 + 3*4; `Cost per pound` := evalf( %/%%, 3) ;

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Problem 1.2 :  You mix 12 pounds of coffee A worth $5.50/lb and 19 pounds of coffee B worth $6.75/lb.

>	MixturePlot( 12, 5.50, 19, 6.75 );    #(a amount) | (a rate) | (b amount) | (b rate)

There are three items we can compute :
        1. How many pounds are in the mixture?
        2. How much does the entire mixture cost?
        3. How much is the mixture worth per pound?

>	`Total number of pounds` := 12 + 19; `Total cost` := 12*5.50 + 19*6.75; `Cost per pound` := evalf( %/%%, 3) ;

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Problem 1.2 :  You mix 12 pounds of coffee A worth $5.50/lb and 19 pounds of coffee B worth $6.75/lb.

Notice the difference between a 50-50 mixture, a 99-1 mixture, and a 1-99 mixture.

>	MixturePlot( 50, 1, 50, 1); MixturePlot( 99, 1, 1, 1 ); MixturePlot( 1, 1, 99, 1 );

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Problem 1.2 :  You mix 12 pounds of coffee A worth $5.50/lb and 19 pounds of coffee B worth $6.75/lb.


There are also other types of mixture problems .....


Problem 1.3 :  You combine 17 nickels and 23 quarters. How many coins to you have? And what is the total value?

>	`number of coins` = 17 + 23; `total value` = 17*.05 + 23*.25;

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Problem 1.4 :  You combine 12 gallons of a 15% acid solution with 34 gallons of a 22% acid solution.
  

2. Mixture Problems                (only one known quantity)

Problem 2.1 :  How many pounds of A nuts worth $2/lb should be combined with 3 pounds of B nuts worth $4/lb to get a mixture worth $3.50 per pound?

Just for fun, we can guess a few answers to get a feel for what is going on. We know all the entries except the quantity of the A nuts. Lets guess 3, 5 and 12 pounds of A to see what price per pound comes out for the result.

>	MixtureTableTest( 3, 2, 3, 4);   # (a quantity)  | (a rate) | (b quantity) | (b rate) MixtureTableTest( 5, 2, 3, 4); MixtureTableTest( 12, 2, 3, 4);

Look at the A+B rate in each case ... $3/lb,   $11/4 = $2.75 /lb, and $12/5 = $2.40/lb.




Lets find the solution. We form the table, form an equation by adding the first two entries in the last column to be equal to the last entry, then solve to find the answer.

>	MixtureTable1( x, 2, 3, 4, 3.5);     #||a amount|a rate|b amount|b rate|c rate||

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Problem 2.2 :  How many kilograms of candy worth $6.50kg should be combined with 12 kg of candy worth $11/kg to get a mixture worth $8 per kg?

>	MixtureTable1( x, 6.50, 12, 11, 8);

Problem 2.3 :   100 grams of chemicals worth $30/g should be mixed with how many grams of a chemical worth $44/gram to obtain a mixture worth $38/gram.

>	MixtureTable1( 100, 30, x, 44, 38);

3. Mixture Problems              (both quantities unknown)

    Coffee & Rice Problems

Problem 3.1 :  How many pounds of coffee A worth $8/lb and coffee B worth $7.25/lb should be mixed to obtain 50 pounds of a mixture worth $7.70 per pound?

We create a table. Since both quantities (of A and B) are unknown, we need to make one x, and the other 50-x. Then we multiply across, add down, to form an equation. In this case : (8x) + (7.25 - 362.50) = 385. Then we solve and enjoy.

>	MixtureTableEq( 8, 7.25, 7.70, 50);      #||a rate|b rate|c rate|c amount||

Here is a view of the entire solution.

>	MixtureTableSoln( 8, 7.25, 7.70, 50 );  #||a rate|b rate|c rate|c amount||

Problem 3.2 :  How many pounds of basmati rice worth $2.70 cents per kg, and wild rice worth $4.57 per kg  should be mixed to obtain 22 kg of a mixture worth $3.55 per kg? Again, we create a table. Since both quantities of rice are unknown, we need to make one x, and the other 22-x. Then we "multiply across" and "add down" to form an equation.

>	MixtureTableEq( 2.70, 4.57, 3.55, 22);      #||a rate|b rate|c rate|c amount||

We solve the equation  : -1.87*x + 100.54 = 78.10, to get x, which is the quantity of A (Basmati rice). We find that A is 12 kg. To find B (Wild rice), we subtract this from 22 to get 10 kg.

Here is a view of the entire solution.

>	MixtureTableSoln( 2.60, 4.20, 3.15, 18 );  #||a rate|b rate|c rate|c amount||

     Coin Problems

Coin problems are similar to coffee, candy, and other "price per pound" quantites. In the case of coins, we don't give them a value per pound, we give them a value per coin. So instead of "number of pounds" and "price per pound", we use "number of coins" and "value per coin". The rest is then the same, essentially.

Problem 3.3 :  How many pennies and quarters are needed to have 64 coins worth $4.72?

Note that both the number of pennies and quarters are both unknown. (Note - not for doing the math, but for using this maple procedure. The "average value" of a coin is the total value of all coins divided by the total number of coins.)

>	MixtureTableEq( .01,.25, 4.72/64 , 64);      #||a rate|b rate|c rate|c amount||

Problem 3.4 :  How many dimes and nickels are needed to have 80 coins worth $4.35?

>	MixtureTableEq( .10,.05, 4.35/80 , 80);

>