28. lawnco produces three grades of commercial fertilizers. a 100 - lb bag of grade a fertilizer contains 18…

28. lawnco produces three grades of commercial fertilizers. a 100 - lb bag of grade a fertilizer contains 18 lb of nitrogen, 4 lb of phosphate, and 5 lb of potassium. a 100 - lb bag of grade b fertilizer contains 20 lb of nitrogen, 4 lb each of phosphate and potassium. a 100 - lb bag of grade c fertilizer contains 24 lb of nitrogen, 3 lb of phosphate, and 6 lb of potassium. formulate and find how many 100 lb bags of each of the three grades of fertilizers should lawnco produce if 26,400 lb of nitrogen, 4900 lb of phosphate, and 6200 lb of potassium are available and all the nutrients are used?

28. lawnco produces three grades of commercial fertilizers. a 100 - lb bag of grade a fertilizer contains 18 lb of nitrogen, 4 lb of phosphate, and 5 lb of potassium. a 100 - lb bag of grade b fertilizer contains 20 lb of nitrogen, 4 lb each of phosphate and potassium. a 100 - lb bag of grade c fertilizer contains 24 lb of nitrogen, 3 lb of phosphate, and 6 lb of potassium. formulate and find how many 100 lb bags of each of the three grades of fertilizers should lawnco produce if 26,400 lb of nitrogen, 4900 lb of phosphate, and 6200 lb of potassium are available and all the nutrients are used?

Answer

Explanation:

Step1: Define variables

Let $x$, $y$, and $z$ be the number of 100 - lb bags of grade A, grade B, and grade C fertilizers respectively.

Step2: Set up the system of equations

Based on the nitrogen content: $18x + 20y+24z=26400$ (divide by 2) $\Rightarrow9x + 10y+12z = 13200$. Based on the phosphate content: $4x + 4y+3z=4900$ (multiply by 4) $\Rightarrow16x + 16y+12z=19600$. Based on the potassium content: $5x + 4y+6z=6200$ (multiply by 2) $\Rightarrow10x + 8y+12z=12400$.

Step3: Eliminate $z$

Subtract the first - nitrogen equation from the second - phosphate equation: $(16x + 16y+12z)-(9x + 10y+12z)=19600 - 13200$ $7x+6y = 6400$. Subtract the first - nitrogen equation from the third - potassium equation: $(10x + 8y+12z)-(9x + 10y+12z)=12400 - 13200$ $x - 2y=-800$, then $x = 2y - 800$.

Step4: Substitute $x$ into $7x+6y = 6400$

$7(2y - 800)+6y=6400$. $14y-5600 + 6y=6400$. $20y=6400 + 5600$. $20y=12000$. $y = 600$.

Step5: Find $x$

Substitute $y = 600$ into $x = 2y - 800$. $x=2\times600 - 800$. $x = 400$.

Step6: Find $z$

Substitute $x = 400$ and $y = 600$ into $4x + 4y+3z=4900$. $4\times400+4\times600+3z=4900$. $1600 + 2400+3z=4900$. $4000+3z=4900$. $3z=900$. $z = 300$.

Answer:

400 bags of grade A, 600 bags of grade B, 300 bags of grade C.