00:01
Hello students, so let us look at this question.
00:05
So profit p of q can be calculated as r of q minus c of q and r of q is given as r of q is given as minus 0 .002 q raised to the power 2 plus 40 q and c of q is given as 0 .0005 q raised to the power 2 plus 10 q plus 60 ,000.
00:39
Now we can substitute this equation into the profit equation that is above one, this one and we can get the desired answer.
00:49
So p of q is equal to minus 0 .005 q minus 0 minus 0 .0025 q raised to the power 2 plus 30 q minus 60 ,000.
01:12
Now to maximize the profit, we can take the derivative of pq with respect to q and we can set it equal to 0.
01:23
So derivative of pq would be equal to minus 0 .005 q plus 30 and after solving equal to 0, we will get q value as 6000.
01:38
Now therefore the production level q that maximizes the profit is 6000 units.
01:45
Now coming on to the second part, the cost elasticity that is el or ei and when q is equal to 1000 can be calculated as ei is equal to dc by c divided by dq by q.
02:11
So the thing which are given to as c of q which is equal to 0 .0005 q raised to the power 2 plus 10 q plus 60 ,000 and q value is given as 1000.
02:28
So let's find the derivative of c of q with respect to the q...