Suppose the marginal revenue (in dollars) from the sale of x jet planes is R'(x)=2x(x^2+50)^2
A. Find the total revenue function if the revenue from 3 planes is $206,379.
B. How many planes must be sold for revenue of at least $450,000?
A. To find the total revenue function, we need to integrate the marginal revenue function with respect to x. The integration will give us the revenue function.
∫ R'(x) dx = ∫ 2x(x^2+50)^2 dx
To solve this integral, we can use the power rule for integration:
∫ x^(n) dx = (x^(n+1))/(n+1)
Applying this rule, we get:
∫ 2x(x^2+50)^2 dx = (2/3)((x^2+50)^3) + C
Since we are given that the revenue from 3 planes is $206,379, we can use this information to find the constant of integration, C. Let's substitute x = 3 and R(x) = 206379 into the equation:
206379 = (2/3)((3^2+50)^3) + C
Now we can solve for C:
206379 = (2/3)(53^3) + C
206379 = (2/3)(148877) + C
206379 = 99317.33 + C
206379 - 99317.33 = C
107061.67 = C
So the total revenue function is:
R(x) = (2/3)((x^2+50)^3) + 107061.67
B. To find the number of planes that must be sold for a revenue of at least $450,000, we can set up an inequality using the revenue function:
R(x) ≥ 450000
Substituting the revenue function we found earlier, we get:
(2/3)((x^2+50)^3) + 107061.67 ≥ 450000
To solve this inequality, we can subtract 107062.67 from both sides to isolate the function:
(2/3)((x^2+50)^3) ≥ 450000 - 107061.67
(2/3)((x^2+50)^3) ≥ 342938.33
Now, we can multiply both sides by 3/2 to get rid of the fraction:
(x^2+50)^3 ≥ (342938.33)(3/2)
Taking the cube root of both sides to isolate the term (x^2+50), we get:
x^2 + 50 ≥ (342938.33)(3/2)^(1/3)
Now, subtracting 50 from both sides gives us:
x^2 ≥ (342938.33)(3/2)^(1/3) - 50
Taking the square root of both sides, we find:
x ≥ sqrt((342938.33)(3/2)^(1/3) - 50)
So, the minimum number of planes that must be sold for a revenue of at least $450,000 is approximately equal to sqrt((342938.33)(3/2)^(1/3) - 50).