00:01
In this problem we are as to find the level of nitrogen that gives the best yield.
00:10
The function of nitrogen is given as y and it is equal to k times n divided by 25 plus n square, where the value of k is a positive integer.
00:27
Now, to find the best yield of nitrogen, we have to find the value of y -dash and we need to equate this to 0 to find the value of corresponding n.
00:41
So let's find the value of y dash from y.
00:45
That is we need to differentiate y with respect to the variable n.
00:50
To differentiate the rhs of the given expression, we are going to use u by v method.
00:57
Whenever we are given a function which is of the form u divided by v and we are as to find the derivative of this expression with respect to x, this will be equal to v times u dash minus u times v dash divided by v square in our case the value of u is equal to k times n and the value of v is equal to 25 plus n square so the value of u dash will be equal to k and the value of v dash will be equal to two times n so the value of v square will be equal to 2 times n to 25 plus n square the whole square.
01:41
So let's substitute all the values in this expression.
01:44
So we will have y dash to be equal to 25 plus n square times k minus k n times 2n divided by 25 plus n square the whole square...