# Mixture Problems

You want to make a mixture of peanuts and cashews to sell. Peanuts cost \$3 per pound and cashews cost \$9 per pound.

 If you buy 5 pounds of peanuts, you will have to pay $5(\$3)=\$15$. How much do you have to pay for 5 pounds of cashews? \$ If you buy and mix all of these peanuts and cashews, how much money will you have spent on the mixture? \$
 How many pounds does this mixture weigh (5 pounds of peanuts plus 5 pounds of cashews)? pounds
 What is the cost per pound of this mixture (that is, the price you paid divided by the weight of the mixture)? \$per pound Let’s see what happens to the cost of a 10-pound mixture if you mix different amounts of peanuts and cashews.  Peanuts cost \$3 per pound. What is the cost of $x$ pounds of peanuts? \$ If a mixture costs$y$dollars per pound, what is the cost of 10 pounds of the mixture? \$

We want to make a 10-pound mixture of peanuts and cashews. This means that if we use $x$ pounds of peanuts, we will have to use $10-x$ pounds of cashews. Cashews cost \$9 per pound, so the total cost (in dollars) of the cashews will be$9(10-x)$. The total cost of the mixture can be found by adding the cost of the cashews to the cost of the peanuts. That is,$\text"Total Cost" = 3x + 9(10-x)$. If$y$is the cost per pound of the mixture, then we have: $$10y=3x+9(10-x)$$  On the grid to the left you can see the graph of$10y=3x+9(10-x)$. Look to the right of the graph. Under the words Trace at: you see$x=5$and$y=6$. This means that when you use 5 pounds of peanuts in your mixture, the cost of the mixture is \$6 per pound. Is this the same as your final answer in mixExample?
 Click or tap inside the grid to the left and drag to the left or right. Move the vertical bar so that $x$ has the value 6.0. This means that you are using 6 pounds of peanuts in your mixture. What is the value of $y$ when $x=6.0$ (that is, how many dollars per pound does the mixture cost now)?
 The mixture is 10 pounds, so if you use 6 pounds of peanuts, how many pounds of cashews are you using? pounds
 Move the vertical bar to find the price per pound of the mixture when you use 2 pounds of peanuts (and 8 pounds of cashews). \$per pound  As you use more peanuts, does the price of the mixture increase or decrease?  As you use fewer peanuts (i.e. more cashews), does the price of the mixture increase or decrease?  To keep your cost down, should you use more peanuts or cashews? You want to combine a 20% methanol solution with a 60% methanol solution to make 10 liters of a mixture.  If you use 5 liters of the 20% solution and 5 liters of the 60% solution, what will the percentage of methanol (the concentration) be in the mixture? %  Rewrite your answer to the previous question as a decimal.  On the grid to the left,$x$is the number of liters of the 20% solution and$y$is the concentration of the mixture, so this time the equation is$10y=0.2x+0.6(10-x)$. What is the concentration of the mixture (the value of$y$) when$x=5$?  Does this match your answer to the previous question?  Use the mouse or touch to move the vertical bar on the grid to find the concentration of the mixture if you use 7 liters of the 20% solution (i.e. find$y$when$x=7$).  As you use more of the 20% solution, does the concentration of the mixture increase or decrease?  How much of each solution should be used to make 10 liters of a 52% solution? liters 20% solution liters 60% solution  Can you make a 70% solution? Another way to look at the 52% solution part of alc1 is as a system of equations. We want to know how much of each of the 20% and 60% solutions should be mixed to get 10 liters of a 52% solution. This time, let$x$be the number of liters of the 20% solution and$y$be the number of liters of the 60% solution. Then the system of linear equations is: $$\{\,\cl"tight"{\table 0.20x,+,0.60y,=,10(0.52); x,+,y,=,10}$$ The first equation in this system represents the total amount of methanol in the mixture, while the second equation represents the total amount of liquid.  This system is graphed to the left. Drag the vertical bar on the grid to find the point of intersection of the two lines. How much of each solution should be used? liters 20% solution liters 60% solution  Does your answer match your answer to the 52% solution part of alc1? In alc2, you found out how much of the 20% and 60% solutions need to be mixed to make 10 liters of a 52% solution. Now let’s mix solutions with different concentrations and see how much of each we need to make 10 liters of a 52% solution. Call the concentration of the first solution$a$and the concentration of the second solution$b$. Using$a$and$b$, our system becomes: $$\{\,\cl"tight"{\table ax,+,by,=,10(0.52); x,+,y,=,10}$$ Using the sliders, complete the table below.$x$is the amount of the first solution and$y$is the amount of the second solution. Remember to use the vertical bar in the grid to the left to find the point of intersection of the two lines.$ab$liters of 1st solution liters of 2nd solution  Set$b$to 0.52. Does changing$a$affect the amount of each solution you need to make 10 liters of a 52% solution?  Set$a$to 0.60 and$b$to 0.70, so that you are mixing a 60% solution and a 70% solution. Do the two lines intersect anywhere in the first quadrant (that is, anywhere with$x≥0$and$y≥0\$)?
 Is it possible to make a 52% solution by mixing a 60% solution and a 70% solution?