Derivative

If I remember correctly, extreme points (max or min) happen where the derivative is zero. Derivative represents the slope of the tangent line, which would be zero (horizontal) at extreme points. So you want to take the derivative (R bar(x)) and set it to zero. There should be two solutions. You can plug each into the original equation and one of them should be the max.

As for b, I don't know what MR represents so I'm not sure what you need to do there.

to get an average you need the range of x

this is done by integrating the function between x= a & b and dividing this integral by b-a.

The function maximum is determined from the derivative

iff'n R(x) = 2800x+8x^2-x^3

then dR(x)/dx=2800+16x-3x^2

set this equal to zero and solve for x

2800+16x-3x^2=0

x=[16+/-sqrt(16^2+4*3*28000]/(2*3)=(16+/-184)/6

a negative x makes no sense so

x=(16+184)/6=33.3333

check to see iff'n this is a max and take the 2nd derivative

d^2R(x)/dx^2=16-6x and let x=33.3333

yep it's negatory so it's a max

so the max revenue is at x=33.3333

Rap

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