# 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?

8 months ago

## 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.

âˆ« R'(x) dx = âˆ« 2x(x^2+50)^2 dx

To solve this integral, we can use the power rule for integration:

âˆ« x^(n) dx = (x^(n+1))/(n+1)

Applying this rule, we get:

âˆ« 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) â‰¥ 450000

Substituting the revenue function we found earlier, we get:

(2/3)((x^2+50)^3) + 107061.67 â‰¥ 450000

To solve this inequality, we can subtract 107062.67 from both sides to isolate the function:

(2/3)((x^2+50)^3) â‰¥ 450000 - 107061.67
(2/3)((x^2+50)^3) â‰¥ 342938.33

Now, we can multiply both sides by 3/2 to get rid of the fraction:

(x^2+50)^3 â‰¥ (342938.33)(3/2)

Taking the cube root of both sides to isolate the term (x^2+50), we get:

x^2 + 50 â‰¥ (342938.33)(3/2)^(1/3)

Now, subtracting 50 from both sides gives us:

x^2 â‰¥ (342938.33)(3/2)^(1/3) - 50

Taking the square root of both sides, we find:

x â‰¥ 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).

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