The management of AB, producers of the famous hot sauce, estimate that their
profit (in RM) from the daily production and sales of x cases (each case
consisting of 24 bottles) of the hot sauce given by
𝑃(𝑥) = −0.000002𝑥
3 + 6𝑥 − 400
What is the largest possible profit AB can make in 1 day?
plz I have to to send it as fast as I can
To find the largest possible profit AB can make in one day, we need to determine the maximum value of the profit function 𝑃(𝑥) = −0.000002𝑥^3 + 6𝑥 − 400.
To find the maximum value, we need to find the critical points of the function. This can be done by taking the derivative of the function and setting it equal to zero.
Let's differentiate 𝑃(𝑥) with respect to 𝑥:
𝑃'(𝑥) = -0.000006𝑥^2 + 6
Setting 𝑃'(𝑥) equal to zero and solving for 𝑥:
-0.000006𝑥^2 + 6 = 0
0.000006𝑥^2 = 6
𝑥^2 = 6 / 0.000006
𝑥^2 = 1,000,000
𝑥 = √1,000,000
𝑥 = 1000
So, 𝑃'(𝑥) is equal to zero when 𝑥 = 1000.
Next, we need to determine if this critical point is a maximum or minimum by evaluating the second derivative of 𝑃(𝑥).
Taking the second derivative of 𝑃(𝑥):
𝑃''(𝑥) = -0.000012𝑥
Since 𝑃''(𝑥) is negative for all 𝑥, the critical point 𝑥 = 1000 corresponds to a maximum.
Now, let's substitute 𝑥 = 1000 into the profit function to find the largest possible profit AB can make in one day.
𝑃(1000) = -0.000002(1000)^3 + 6(1000) - 400
𝑃(1000) = -0.000002(1,000,000,000) + 6,000 - 400
𝑃(1000) = -2,000 + 6,000 - 400
𝑃(1000) = 3,600
Therefore, the largest possible profit AB can make in one day is RM 3,600.
To find the largest possible profit, we need to determine the maximum value of the profit function P(x) = -0.000002x^3 + 6x - 400.
To do this, we can calculate the derivative of the profit function and find the critical points. The critical points are the values of x at which the derivative equals zero or is undefined.
Let's find the derivative of P(x):
P'(x) = -0.000006x^2 + 6
Now, let's set P'(x) = 0 and solve for x:
-0.000006x^2 + 6 = 0
Dividing by -0.000006:
x^2 = 1000000
Taking the square root of both sides:
x = ±1000
Since the number of cases cannot be negative, we discard the negative value:
x = 1000
Thus, x = 1000 is the only critical point.
Next, we need to check whether this critical point corresponds to a maximum or a minimum. To do this, we can use the second derivative test.
Let's find the second derivative of P(x):
P''(x) = -0.000012x
Now, let's evaluate P''(1000):
P''(1000) = -0.000012 * 1000 = -0.012
Since the second derivative is negative, this indicates that the critical point x = 1000 corresponds to a maximum value.
Finally, we can find the largest possible profit by plugging x = 1000 into the profit function:
P(1000) = -0.000002 * 1000^3 + 6 * 1000 - 400
P(1000) = -0.000002 * 1000000000 + 6000 - 400
P(1000) = -2000 + 6000 - 400
P(1000) = 3600
Therefore, the largest possible profit AB can make in 1 day is RM 3600.
I cannot interpret P(x)
what does the 2nd line mean? Is it a rational function, or do you mean x^3 ?
If 𝑃(𝑥) = −0.000002𝑥^3 + 6𝑥 − 400
then P'(x)=0 tells you that P(x) has a maximum at (1000,3600)