(16) Suppose the revenue from producing​ (and selling) x units of a product is given by R(x) = 7x − .02x^2 dollars.

Question

​(a) Find the marginal revenue at a production level of 15.



​(b) Find the production levels where the revenue is $300.

Answer 1

(a) The marginal revenue is the derivative of the revenue function. Using the power rule, R'(x) = 7 - 0.04x

So R'(15) = 7 - 0.04(15) = $6.4/unit

(b) Since R(x) = 7x - 0.02x², when the revenue is 300 we can say:

300 = 7x - 0.02x²

-0.02x² + 7x - 300 = 0

Using the quadratic formula, for ax² + bx + c = 0

So, x = [-7 ± √(7² - 4(-0.02)(-300))]/(2(-0.02))

x = (-7 ± √25)/(-0.04)

x = (-7 + 5)/(-0.04) and x = (-7 - 5)/(-0.04)

x = 50 units and x = 300 units