00:01
Pounds of nitrogen and p is pounds of phosphate.
00:03
And why is the yield in bushels of corn? so how much corn will be produced with respect to how much nitrogen and phosphate you put on the field? this farmer has $30 an acre to spend on nitrogen and phosphate.
00:22
And nitrogen is 25 cents a pound.
00:25
And phosphate is 20 cents a pound.
00:27
Find n &p to maximize z.
00:30
Yield.
00:31
So we want to maximize y.
00:34
There's a bunch of decimals in there.
00:36
So i'm going to maximize a thousand y, which if we find the maximum of a thousand y, it will be the same as finding the maximum of y.
00:46
So that's 7500 plus 600, oh 600 n plus 700 p, minus n squared minus 2 p squared plus np.
01:01
Okay and that's subject to the constraint that he has $30 to spend and cost 25 cents a pound for n and 20 cents a pound for p.
01:27
Okay, so we're going to take the derivative of 1 ,000 y and set it equal to zero and solve.
01:32
But first we've got to get the n or p out of there and it really doesn't matter which one you solve for.
01:38
I'm going to solve for p, i guess.
01:42
No, n.
01:43
I'm going to solve for n.
01:47
Okay, so 30.
01:49
Ops.
01:52
0 .25n equals 30 minus 0 .2 o .p.
01:59
So n equals 30 minus 0 .2op, divide by 0 .25, the whole thing.
02:06
That was bad writing there.
02:07
So that's equal to 120 minus 0 .8p.
02:15
All right.
02:16
So now i'm going to take this 1 ,000 y.
02:18
I'm just going to call it big y now, fancy y...