Problem 35 - Calculus for Biology and Medicine

Question

SplendidLawn sells three types of lawn fertilizer: SL 24-4- 8, SL 21-7-12 and SL \(17-0-0 .\) The three numbers refer to the percentages of nitrogen, phosphate, and potassium, in that order, of the contents. (For instance, \(100 \mathrm{~g}\) of SL 24-4-8 contains \(24 \mathrm{~g}\) of nitrogen.) Suppose that each year your lawn requires \(500 \mathrm{~g}\) of nitrogen, \(100 \mathrm{~g}\) of phosphate, and \(180 \mathrm{~g}\) of potassium. How much of each of the three types of fertilizer should you apply?

Step-by-Step Solution

Answer

You should apply 2500 g of SL 24-4-8, 1000 g of SL 21-7-12, and 0 g of SL 17-0-0.

Step 1: Define Variables

Let's define the variables representing how much of each fertilizer is used. Let \( x \) be the amount of SL 24-4-8, \( y \) be the amount of SL 21-7-12, and \( z \) be the amount of SL 17-0-0.

Step 2: Create Equations for Nitrogen

Each fertilizer contains nitrogen. From the problem statement:\[ 0.24x + 0.21y + 0.17z = 500 \]This equation ensures that the total nitrogen content meets the requirement of 500 g.

Step 3: Create Equations for Phosphate

The phosphate content must also meet requirements:\[ 0.04x + 0.07y + 0 = 100 \]Here, only SL 24-4-8 and SL 21-7-12 contribute phosphate.

Step 4: Create Equations for Potassium

Similarly, for the potassium content:\[ 0.08x + 0.12y + 0 = 180 \]The contribution in potassium comes from SL 24-4-8 and SL 21-7-12.

Step 5: Solve the System of Equations

Now we solve the system of three equations:1. \( 0.24x + 0.21y + 0.17z = 500 \)2. \( 0.04x + 0.07y = 100 \)3. \( 0.08x + 0.12y = 180 \)Use methods such as substitution or elimination to solve these equations. After computation, we find:\(x = 2500\) g,\(y = 1000\) g,and \(z = 0\) g.

Step 6: Verify the Solution

Double-check the solution by inserting the values of \(x\), \(y\), and \(z\) back into the original three equations to ensure all nutrient requirements are satisfied.

Key Concepts

Linear Algebra

Linear algebra involves the study of vectors, matrices, and systems of linear equations. When working with systems of equations in this context, we seek to find the values of the variables at play. For the problem at hand, the fertilizers' nutrient content is represented through a system of linear equations.
Each equation corresponds to a nutrient requirement—nitrogen, phosphate, and potassium. The coefficients of the variables reflect the percentage of each nutrient in each fertilizer type. This forms a linear combination that must equal the nutritional needs of the lawn. Understanding these relationships can demystify the task of deciding how much of each fertilizer is necessary.
Linear algebra provides a systematic way to tackle such problems, offering clarity when faced with potentially complex systems. It reduces them to a manageable set of operations, making practical applications like nutrient analysis accessible to everyone.

Nutrient Analysis

Nutrient analysis in fertilizers helps determine the optimal blend to meet specific requirements. In this problem, we need to ensure the lawn receives precise quantities of nitrogen, phosphate, and potassium. By analyzing these nutrients in the available fertilizers, it's possible to deduce the correct mix.
The percentages of each nutrient in the fertilizers (SL 24-4-8, SL 21-7-12, and SL 17-0-0) dictate how much of each fertilizer needs to be used to achieve the desired nutritional content of the lawn. For example:

• SL 24-4-8: 24% nitrogen, 4% phosphate, 8% potassium.
• SL 21-7-12: 21% nitrogen, 7% phosphate, 12% potassium.
• SL 17-0-0: 17% nitrogen, 0% phosphate, 0% potassium.

With this analysis, it becomes a straightforward calculation to match the required nutrient amounts with the most efficient use of the fertilizers provided.

Substitution and Elimination Method

The substitution and elimination methods are handy techniques for solving systems of equations, involving multiple unknowns like in our fertilizer problem. The goal is to isolate variables, transforming the system into simpler equations.
**Substitution Method**: This involves solving one of the equations for one variable and then substituting this expression into the other equations. It simplifies the problem gradually by reducing the number of variables in other equations.
**Elimination Method**: This technique focuses on eliminating a variable altogether by adding or subtracting equations. We adjust the coefficients so that when two equations are added or subtracted, one of the variables cancels out. This reduces the system from three variables to two, making the solution easier to find.
In our fertilizer example, breaking down the three original equations into manageable expressions ultimately allows for finding the exact quantities of fertilizer components needed. Mastering these methods provides flexibility and accuracy in handling many types of algebraic challenges.