Wheat Farming Problem Derivatives I am currently learning derrivatvies in our class and i am faced with this problem at hand.

Wheat farmer is converting to corn because he believes that corn is a
  more lucrative crop. It is not feasible for him to convert all his
  acreage to corn at once. He is farming 500 acres of corn in the
  current year and is increasing that number by 50 acres per year. As he
  becomes more experienced in growing corn, his output increases. He
  currently harvests 130 bushels of corn by acre, but the yield is
  increasing by 5 bushels per acre per ear. When both the increasing
  acreage and the increasing yield are considered, how rapidly is the
  total number of bushels of corn increasing per year.

Well what i did was i know the next year he has $130 + 5$ or $135$ bushels per acre since it increases $5$ acres per year. And then he adds $50$ acres per year. So $50 * 135$ should give you the number of bushels of corn increasing per year.. Which is $6750$ bushels per year.
But when i look on answer key on my text book the answer is $9000$ bushels per year. Why is that?
 A: Let $f(n)$ denote the number of acres devoted to corn in year $n$.  We have $$f(n)=500+50n$$
Let $g(n)$ denote the yield, per acre, in year $n$.  We have $$g(n)=130 +5n$$
It follows that the $t(n)$, the number of bushels of corn harvested in year $n$, is $$t(n)=f(n)\times g(n)=\left(500+50n\right)\times \left(135+5n\right)$$
We deduce that $$t'(n)=f'(n)g(n)+f(n)g'(n)=50g(n)+5f(n)$$
Thus $$t'(0)=50\times 130+5\times 500=9000$$
A: Units are always important because they tell you if your reasoning has been correct.
The number of acres is: 
$$ a(y) = 500 \text{acres}  + 50\text{(acres / year)}\times y, $$
where $y$ is the number of years. Clearly $y$ is measured in years, and thus $a$ is measured in acres, as it should!
The total number of bushes is:
$$  b(a, y) = [130 \text{(bushels / acre)} + 5 \text{(bushels / (acre*year))}\times y]\times a.  $$
We see that the equations are correct because $b$ is measured in bushels. Continuing we drop the units and we end up with:
$$b(y) = (130 + 5y)\overbrace{(500 + 50y)}^{a(y)}.$$
From here you derive and evaluate in the present, i.e. $y=0$. However the question should be: how rapidly is the total number of bushels of corn increasing per year currently?
