00:01
Hello, so we know that the revenue is the product of the price and the quantities.
00:04
We have that revenue is equal to price times quantity.
00:10
So we're given here, the constraints given to us by p is equal to 6 ,570 ,000 over q to the 1 .3.
00:18
We know that q is greater than equal to 10 ,000.
00:21
So we can eliminate one of our variables and we can restate our problem as rfq is equal to 6 ,700.
00:30
50 ,000 times q to the negative 0 .3, and this is subject to q being greater than or equal to 10 ,000.
00:40
Okay, to find our stationary points, then we find the derivative here of this function, and then equate our derivative equal to zero.
00:47
So our derivative is going to be r prime of q, which is going to be equal to while we get 6 ,750 ,000 times negative 0 .3, and then times q to the negative 1 .3.
01:02
So that's going to give us 2 ,025 ,000 over q to the 1 .3.
01:13
So there's our derivative.
01:14
And to find our stationary point, we take our derivative and set this equal to 0.
01:20
But the only way that a fraction can be equal to 0 is if the numerator is equal to 0, and our numerator here is never equal to 0.
01:27
So therefore, we have no stationary points...