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Matrix Question

  1. Oct 8, 2008 #1
    1. The problem statement, all variables and given/known data

    A fertiliser company sells four types of lawn fertiliser.
    Brand A 24-4-8, Brand B 21-7-12, Brand C 17-0-0 , Brand D 0-12-12. The three
    numbers refers to the percentage of nitrogen, phosphate and potassium in that order
    contained in each brand. Suppose that each year your lawn requires 500g of nitrogen,
    100g of phosphate and 180g of potassium. You want to find the amount of each of
    the four brands of fertiliser that is required to do this.
    (a) Set up this information as a matrix equation with the coefficient matrix
    of size 3x4.
    (b) Using row transformations solve for the amount of each brand of fertiliser.
    (c) Find the exact amount for each brand when you use
    (i) the minimum amount of Brand D allowed
    (ii) the maximum amount of brand D allowed


    2. Relevant equations



    3. The attempt at a solution

    I'm having trouble working out how to attach this. We've only really solved for unknowns with neat square matrices before.

    a)

    I throw it together in a matrix like this:

    24, 21, 17, 0 : 500
    4, 7, 0, 12 : 100
    8, 12, 0, 12: 180

    But then i don't know what to do, because the # of unknowns is more than the # of equations.

    Any help?
     
  2. jcsd
  3. Oct 8, 2008 #2

    gabbagabbahey

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    You have 4 unknowns and only 3 equations, so one of your unknowns will be left as a variable parameter; that is, you can choose a value for one brand and the calculate the amounts of the other three. Since part(c) wants you to vary the amount of Brand D, I recommend you solve your matrix in a way that leaves the amount of brand D as a parameter. One way to do this, is to call this parameter [itex]t[/itex], giving you an extra equation, namely [itex]D=t[/itex] and row-reduce the following Matrix:

    [tex] \begin{bmatrix} 24 & 21 & 17 & 0 & 500 \\ 4 & 7 & 0 & 12 & 100 \\ 8 & 12 & 0 & 12 & 180 \\ 0 & 0 & 0 & 1 & t \end{bmatrix}[/tex]
     
  4. Oct 8, 2008 #3
    Oh ok.

    So my amounts of brands A, B, and C, are going to end up being given in terms of t?
     
  5. Oct 8, 2008 #4

    gabbagabbahey

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    Yes, they should be in terms of t.
     
  6. Oct 8, 2008 #5
    Thanks so much for your help.


    I've done as you said and ended up with:
    A = 16 2/3 + (7 1/2)t
    B = 10 - 6t
    C = 6 8/17 - (3 3/17)t
    D = t

    Does this look about what I should have expected?

    Also, any advice as to answer part (c) from this information?
     
  7. Oct 8, 2008 #6

    gabbagabbahey

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    B,C,D are correct, but A is a little off.

    Once you correct A, part (c) shouldn't be too hard. The minimum value of D is clearly zero since it is a free parameter, and the only physical constraints are that A,B,C and D aren't negative (you can't have a negative amount of fertilizer). As you increase the amount of D, the value of t increases. Look at your equations for A,B and C...what is the largest value of t you can use, for which A,B,C are are all still greater than or equal to zero?
     
  8. Oct 8, 2008 #7
    My mistake.

    I think A should be 45 + (7 1/2)t.


    That's what I was thinking for part (c) but wasn't entirely sure. It makes sense now.

    Thanks again for your assistance. You've been extremely helpful.
     
  9. Oct 8, 2008 #8

    gabbagabbahey

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    Mathematica gives A=7.5 +7.5 t so you should probably go over your algebra once more. What do you end up getting for the maximum amount of D?
     
  10. Oct 8, 2008 #9
    Found my error (again).

    My maximum amount for D was 1 2/3.
     
  11. Oct 8, 2008 #10

    gabbagabbahey

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    looks good to me:approve:
     
  12. Oct 8, 2008 #11
    Yay. :smile:

    hey ho let's go.
     
  13. Oct 8, 2008 #12

    gabbagabbahey

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    shoot 'em in the back now:smile:
     
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