Suppose the revenue from producing (and selling) x units of product is given by R(x)=7x-0.01x^2 dollars.
A)find the marginal revenue at a production level of 25.
B)find the production levels where the revenue is $600.
To find the marginal revenue at a production level of 25, we need to find the derivative of the revenue function, R(x), with respect to x and evaluate it at x = 25.
A) Marginal revenue represents the rate at which revenue changes with respect to the quantity of a product produced. Mathematically, it is the derivative of the revenue function, R(x), with respect to x.
To find the derivative of R(x), we differentiate each term separately using the power rule:
R(x) = 7x - 0.01x^2
dR(x)/dx = d(7x)/dx - d(0.01x^2)/dx
= 7 - 0.02x
Now we can evaluate the derivative at x = 25 to find the marginal revenue:
MR(25) = 7 - 0.02(25)
= 7 - 0.5
= 6.5 dollars
Therefore, the marginal revenue at a production level of 25 is $6.50.
B) To find the production levels where the revenue is $600, we need to set the revenue function, R(x), equal to 600 and solve for x:
R(x) = 7x - 0.01x^2
Setting R(x) = 600, we get:
7x - 0.01x^2 = 600
Rearranging the equation, we have:
0.01x^2 - 7x + 600 = 0
Solving this quadratic equation can be done through factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
For the quadratic equation in the form of ax^2 + bx + c = 0, we have:
a = 0.01, b = -7, c = 600
Plugging these values into the quadratic formula, we get:
x = (-(-7) ± sqrt((-7)^2 - 4(0.01)(600))) / (2(0.01))
= (7 ± sqrt(49 + 24)) / 0.02
= (7 ± sqrt(73)) / 0.02
x = (7 + sqrt(73)) / 0.02 or x = (7 - sqrt(73)) / 0.02
Evaluating these expressions will give us the production levels where the revenue is $600.
Using a calculator, we find:
x ≈ 56.99 or x ≈ 43.01
Therefore, the production levels where the revenue is $600 are approximately 56.99 and 43.01 units.