Solve Matrix Equation for Lawn Fertilizer Amounts

  • Thread starter billmccai
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In summary, a fertiliser company sells four types of lawn fertiliser. The three numbers in parentheses refer to the percentage of nitrogen, phosphate and potassium contained in each brand. Each year, your lawn requires 500g of nitrogen, 100g of phosphate and 180g of potassium. To find the amount of each of the four fertilisers required, you solve a matrix equation with the coefficient matrix of size 3x4.
  • #1
billmccai
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Homework Statement



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


Homework Equations





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?
 
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  • #2
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]
 
  • #3
Oh ok.

So my amounts of brands A, B, and C, are going to end up being given in terms of t?
 
  • #4
Yes, they should be in terms of t.
 
  • #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?
 
  • #6
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?
 
  • #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.
 
  • #8
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?
 
  • #9
Found my error (again).

My maximum amount for D was 1 2/3.
 
  • #10
looks good to me:approve:
 
  • #11
Yay. :smile:

hey ho let's go.
 
  • #12
shoot 'em in the back now:smile:
 

1. How do I solve a matrix equation for lawn fertilizer amounts?

To solve a matrix equation for lawn fertilizer amounts, you will need to set up a system of equations that represents the given information. Then, use matrix operations such as multiplication, addition, and subtraction to solve for the unknown variables. It may be helpful to use a calculator or a computer program to perform the calculations.

2. What information do I need to solve a matrix equation for lawn fertilizer amounts?

To solve a matrix equation for lawn fertilizer amounts, you will need to know the cost of each type of fertilizer, the area of your lawn, and the desired nutrient levels for your lawn. This information will be used to set up the system of equations and find the unknown variables.

3. Are there any special techniques for solving a matrix equation for lawn fertilizer amounts?

Yes, there are some special techniques that can be used to solve a matrix equation for lawn fertilizer amounts. These include Gaussian elimination, Cramer's rule, and using inverse matrices. These techniques can make the solving process more efficient and accurate.

4. Can I use a matrix equation to determine the best combination of fertilizers for my lawn?

Yes, a matrix equation can be used to determine the best combination of fertilizers for your lawn. By setting up the system of equations with different combinations of fertilizers, you can find the one that meets your desired nutrient levels at the lowest cost.

5. How do I know if my solution for a matrix equation on lawn fertilizer amounts is correct?

You can check the accuracy of your solution by plugging the values into the original equations and seeing if they satisfy the given information. You can also use a calculator or a computer program to perform the calculations and compare the results to your solution.

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