00:02
Suppose, p is the demand function, a selling price of an item to the quantity x sold, 8x plus 100, where x belongs to 0 less than x less than 12.
00:23
Now what is the maximum revenue possible in this situation and assume that p is the maximum revenue.
00:34
So revenue function first, xp, i'll be multiplying this with x, minus 1 by 8x square plus 100x.
00:43
So for maximum revenue, r 'x will be equal to 0 for critical points.
00:48
So differentiating this, minus 1 by 4x plus 100 equal to 0, x is 400.
00:57
If we need to ascertain if it is a point of maximum or minimum, r''x will be simply minus 1 by 4, which is negative...