Answer:
The maximum revenue is 16000 dollars (at p = 40)
Step-by-step explanation:
One way to find the maximum value is derivatives. The first derivative is used to find where the slope of function will be zero.
Given function is:
[tex]R(p) = -10p^2+800p[/tex]
Taking derivative wrt p
[tex]\frac{d}{dp} (R(p) = \frac{d}{dp} (-10p^2+800p)\\R'(p) = -10 \frac{d}{dp} (p^2) +800 \ frac{d}{dp}(p)\\R'(p) = -10 (2p) +800(1)\\R'(p) = -20p+800\\[/tex]
Now putting R'(p) = 0
[tex]-20p+800 = 0\\-20p = -800\\\frac{-20p}{-20} = \frac{-800}{-20}\\p = 40[/tex]
As p is is positive and the second derivative is -20, the function will have maximum value at p = 40
Putting p=40 in function
[tex]R(40) = -10(40)^2 +800(40)\\= -10(1600) + 32000\\=-16000+32000\\=16000[/tex]
The maximum revenue is 16000 dollars (at p = 40)
