00:01
We have the production level given by q, where q is the number of units of output, l is the number of units of labor, and k is the number of units of capital.
00:12
The profit p is equal to the revenue minus the cost.
00:17
Okay, so the revenue is the selling price per unit, which is eight dollars per unit, multiplied by the number of units sold.
00:29
So i'm going to assume that all of the units produced are sold, so the number of units sold is q.
00:36
Okay, that's the revenue, and we need to subtract the cost.
00:40
The labor costs are two dollars per unit of labor.
00:45
Okay, so that's 2l total for labor, and the capital costs are one dollar per unit of capital, so total of k dollars.
00:58
Okay, so we have this p equals 8q minus 2l minus k.
01:06
So now i'll plug in the expression for q.
01:12
So we have p equals 8 times 2l to the one -half power plus 3k to the one -half power minus 2l minus k.
01:29
Okay, so then p is equal to 16l to the one -half plus 24k to the one -half minus 2l minus k, and we want to maximize the profit.
01:43
Okay, so maximize p.
01:45
So we need to find the critical points of the function p.
01:52
These critical points are where the partial derivative with respect to l is zero, and the partial derivative with respect to k is zero.
02:03
Okay, so let's calculate these partial derivatives.
02:05
We have partial derivative of p with respect to l is equal to one -half times 16 times l to the negative one -half power minus 2, and the partial derivative with respect to k is one -half times 24 times k to the negative one -half power minus 1.
02:30
Okay, so let's set these both equal to zero.
02:34
Okay, so we have zero equals one -half times 16 is just 8, so i'm going to write 8l to the negative one -half minus 2.
02:45
Okay, so then we have that 2 equals 8l to the negative one -half power, so one -fourth equals l to the negative one -half power, and i'm going to raise both sides by that to the negative second power, so i get 16 is equal to l.
03:11
And i also need to set the other partial derivative equal to zero, so zero equals 12k to the negative one -half power minus 1, so one -twelfth is equal to k to the minus one -half power...