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 and 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. How many 100-lb bags of each of the three grades of fertilizers should Lawnco produce if the following is true?A) 38,800 lb of nitrogen, 7,100 lb of phosphate, and 9,200 lb of potassium are available and all the nutrients are used.B) 33,800 lb of nitrogen, 6,500 lb of phosphate, and 8,100 lb of potassium are available and all the nutrients are used

Answer:a)We need to produce 600 100-lb bags of the grade A fertilizerWe need to produce 800 100-lb bags of the grade B fertilizerWe need to produce 500 100-lb bags of the grade C fertilizerb) We need to produce 700 100-lb bags of the grade A fertilizerWe need to produce 850 100-lb bags of the grade B fertilizerWe need to produce 300 100-lb bags of the grade C fertilizerStep-by-step explanation:I am going to say that:x is the number of 100-lb bags of grade A fertilizery is the number of 100-lb bags of grade B fertilizerz is the number of 100-lb bags of grade C fertilizerWe have that: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 and 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.A) 38,800 lb of nitrogen, 7,100 lb of phosphate, and 9,200 lb of potassium are available and all the nutrients are usedThis means that we have to solve the following system:[tex]18x + 20y + 24z = 38800[/tex][tex]4x + 4y + 3z = 7100[/tex][tex]5x + 4y + 6z = 9200[/tex]In the second equation, i am going to write y as a function of x and z, and replacing of the the first and the third equation.[tex]4x + 4y + 3z = 7100[/tex][tex]y = \frac{7100 - 4x - 3z}{4}[/tex]In the first equation[tex]18x + 20\frac{7100 - 4x - 3z}{4} + 24z = 38800[/tex][tex]18x + 5(7100 - 4x - 3x) + 24z = 38800[/tex][tex]18x + 35500 - 20x - 15z + 24z = 38800[/tex][tex]-2x + 9z = 3300[/tex]In the third equation[tex]5x + 4\frac{7100 - 4x - 3z}{4} + 6z = 9200[/tex][tex]5x + 7100 - 4x - 3z + 6z = 9200[/tex][tex]x + 3z = 2100[/tex]Now we have:[tex]-2x + 9z = 3300[/tex][tex]x + 3z = 2100[/tex]Now we multiply the second equation by 2, and add with the first:[tex]2x + 6z = 4200[/tex][tex]-2x + 2x + 9z + 6z = 3300 + 4200[/tex][tex]15z = 7500[/tex][tex]z = 500[/tex]We need to produce 500 100-lb bags of the grade C fertilizer[tex]x + 3z = 2100[/tex][tex]x = 2100 - 3z = 600[/tex]We need to produce 600 100-lb bags of the grade A fertilizer[tex]y = \frac{7100 - 4x - 3z}{4} = \frac{7100 - 4(600) - 3(500)}{4} = 800[/tex]We need to produce 800 100-lb bags of the grade B fertilizerB) 33,800 lb of nitrogen, 6,500 lb of phosphate, and 8,100 lb of potassium are available and all the nutrients are usedThis means that we have to solve the following system:[tex]18x + 20y + 24z = 33800[/tex][tex]4x + 4y + 3z = 6500[/tex][tex]5x + 4y + 6z = 8100[/tex]In the second equation, i am going to write y as a function of x and z, and replacing of the the first and the third equation.[tex]4x + 4y + 3z = 6500[/tex][tex]y = \frac{6500 - 4x - 3z}{4}[/tex]In the first equation[tex]18x + 20\frac{6500 - 4x - 3z}{4} + 24z = 33800[/tex][tex]18x + 5(6500 - 4x - 3x) + 24z = 33800[/tex][tex]18x + 32500 - 20x - 15z + 24z = 33800[/tex][tex]-2x + 9z = 1300[/tex]In the third equation[tex]5x + 4\frac{6500 - 4x - 3z}{4} + 6z = 8100[/tex][tex]5x + 6500 - 4x - 3z + 6z = 8100[/tex][tex]x + 3z = 1600[/tex]Now we have:[tex]-2x + 9z = 1300[/tex][tex]x + 3z = 1600[/tex]Now we multiply the second equation by 2, and add with the first:[tex]2x + 6z = 4200[/tex][tex]-2x + 2x + 9z + 6z = 1300 + 3200[/tex][tex]15z = 4500[/tex][tex]z = 300[/tex]We need to produce 300 100-lb bags of the grade C fertilizer[tex]x + 3z = 1600[/tex][tex]x = 1600 - 3z = 700[/tex]We need to produce 700 100-lb bags of the grade A fertilizer[tex]y = \frac{6500- 4x - 3z}{4} = \frac{7100 - 4(700) - 3(300)}{4} = 850[/tex]We need to produce 850 100-lb bags of the grade B fertilizer