Can someone help please?
1 answer:
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Answer:
30 cows
Step-by-step explanation:
The given parameters are;
v(x) = x³ + 21·x² - 1480·x
Where x = The number of cows
V(x) = The weekly volume (in liters) of milk produced
The number of cows that corresponds to a total production of 1500 liters of milk in a week is given as follows;
1,500 = x³ + 21·x² - 1480·x
x³ + 21·x² - 1480·x = 1,500
x³ + 21·x² - 1480·x - 1,500 = 0
x³ + 21·x² - 1480·x - 1,500 = 0
We observe that x = -1 is a solution of the above equation, as follows;
(-1)³ + 21·(-1)² - 1480·(-1) - 1,500 = 0
Therefore, (x + 1) is a factor of x³ + 21·x² - 1480·x - 1,500, which by long division gives;
<u>x² + 20·x - 1500</u>
(x³ + 21·x² - 1480·x - 1,500)/(x + 1)
x³ + x²
20·x² - 1480·x - 1500
20·x² + 20·x
-1500·x - 1500
-1500·x - 1500
Therefore, the other factors is x² + 20·x - 1500, which gives;;
x² + 20·x - 1500 = (x - 30) × (x + 50)
Therefore;
x³ + 21·x² - 1480·x - 1,500 = 0
Which gives;
(x - 30) × (x + 50) × (x + 1) = 0
The solutions are x = 30, or x = -50, or x = -1
Therefore, the only possible solution is the number of cows = x = 30 cows.