Find the marginal revenue at x=10 using the equation R(x)=-x^2+80x
Is the answer 60?
R(x) = -x^2 + 80x
R'(x) = -2x + 80
R(10) = -2(10) + 80 = -20 + 80 = 60
Correct
Yes! Thank you! I do have another question, though. The second part of the problem says to "compare R(11) -R(10) with that result." Does that mean that I derive it before plugging in 11 and 10 or...?
no, R(11) is just that. No derivative.
It's showing you how the marginal revenue can be approximated by noting the increase in revenue from one point to another.
R(11) = 759
R(10) = 700
∆R = 59, which is pretty close to R'(10)
To find the marginal revenue, we need to find the derivative of the revenue function with respect to x.
The revenue function is given as R(x) = -x^2 + 80x.
To find the derivative, we apply the power rule for differentiation. For a polynomial of the form ax^n, the derivative is given by bringing down the exponent n as the coefficient and then subtracting 1 from the exponent.
So, the derivative of R(x) = -x^2 + 80x is:
R'(x) = -2x + 80.
Now, to find the marginal revenue at x = 10, we substitute x = 10 into the derivative.
R'(10) = -2(10) + 80
= -20 + 80
= 60.
Therefore, the marginal revenue at x = 10 is indeed 60, not 60.