I have a word problem that reads as so
A farmer wants to fence a rectangular part of his land (30,000 square feet). The fenced area is to have one border shared with a neighbor which he wishes to be fancy so he will spend 25 dollars per linear foot on that side where as he will use 5 dollars per linear foot of fencing on the rest. Find the dimensions that will minimize the cost
I set up my equations (x shares the border with the fancy fence)
$ A = xy $
$ C = 25x + 5(x + 2y) = 30x + 10y $
Then I found my x-value
$ x = \frac{30,000}{y} $
Plugged it in
$ C = 30(\frac{30,000}{y}) + 10y = \frac{900,000}{y} + 10y $
Derivative of C
$ C' = \frac{900,000}{y^2} + 10 $
$ -900,000 = -10y^2 $
$ y^2 = 90,000 $
$ y = 300 $
Finally, plugged it into original equation
$ 30,000 = 300x = x = 100 $
I just wanted to check if I've done this right. If not, can you explain where I've gone wrong?
Best Answer
You write:
Take a good look at that: it says (among other things) that $90,000=300$. Do you really want to say that?
Note too that you’re trying to find an extreme value of something, so you should expect to take a derivative of something.
What you should get from $C=30\left(\frac{30,000}y\right)+10y$ is that $C=\frac{900,000}y+10y$. Now you want to find the value of $y$ that minimizes $C$, so you calculate $\frac{dC}{dy}$, set it to $0$, and solve for critical points. We have
$$\frac{dC}{dy}=-\frac{900,000}{y^2}+10\;,$$
so we set $$-\frac{900,000}{y^2}+10=0$$ and solve: $10y^2=900,000$, $y^2=90,000$, and $y=300$. You still have to check that this critical value yields a minimum of $C$, but that’s easy to do: by inspection $\frac{dC}{dy}$ is negative when $y$ is a little less than $300$ and positive when $y$ is greater than $300$.