00:01
In this problem we have given the revenue function for a one product firm is given by the following equation.
00:08
The revenue function rx is given by 100 minus 1 ,600 by x plus 6 minus x.
00:20
And here we have to find the value of x that results in maximum revenue.
00:26
So x that gives maximum revenue.
00:33
It means we have to find the value of x which maximize this function r x to maximize this function r x first we have to find critical numbers so the critical numbers can be find by equating this derivative of the function r x with respect to x to 0 so first we will find r -dash -x, that is d by d x of rx, 100 minus 16 by x plus 6 minus x.
01:16
So this is equal to 0 minus and 1600 and derivative of 1 by x plus 6 is minus 1 by x plus 6 is and derivative of x is minus 1.
01:32
So r -dash -x is here, 600 by x plus 6 whole square minus 1.
01:42
So to find the critical numbers, we will equate this r -dash x to 0.
01:47
So this is equal to 1600 by x plus 6 whole square minus 1 equal to 0.
01:55
From here we have x plus 6 whole square equal 1600 so x plus 6 can be written as plus minus square root 1 ,600 from here we have x plus 6 equal plus minus 40 so taking positive sign we will have x plus 6 equal 14 40 that is x equal 34 and if you take minus 40, so we will have x plus 6 equal minus 40, that is x equal minus 46...