Finding the absolute minimum

Question

I was given the following problem:

A college is planning to construct a new parking lot. The parking lot must be rectangular and enclose 6,000 square meters of land. A fence will surround the parking lot, and another fence parallel to one of the sides will divide the parking lot into two sections. What are the dimensions, in meters, of the rectangular lot that will use the least amount of fencing?

I understand that I need to find the absolute minimum. I made two functions: $xy=6000$ and $2x+3y=P$. I solved for $y$ in the first problem and got $y=\frac{6000}x$. I then plugged that into my perimeter function: $2x+\frac{18000}x=P$.
Now, I took the derivative of this function: $2-\frac{18000}{x^2}$. I now need to set it to zero. I ended up with an absolute minimum of $x=30$ and $y=200$. But, this does not seem to be right -right now I have the dividing fence as 200 when it can really be only 30. What did I do wrong? How should I solve this problem?

Answer 1

You made a computational error. Actually, $x=30\sqrt{10}$ and $y=20\sqrt{10}$.

Answer 2

If you had done $P = 3x + 2y$ instead, you would get $P = 3x + 2 \times \frac{6000}{x}$. Differentiating $P$ wrt $x$ and setting the expression to 0, we get $3 -\frac{ 12000}{x^2} = 0$. Solving, you get $x = \sqrt {4000} = 20 \sqrt {10}$, and then $y = 30\sqrt {10}$. The way you did it, the answers should have been $x = 30 \sqrt {10}$ and $y = 20\sqrt {10}$. So both ways should give the same minimum $P$.

Answer 3

You just made an arithmetic mistake .

Note that $$2-\frac{18000}{x^2}=0 \implies x^2=9000 \implies x=30\sqrt {10}$$ and then you find your $$y= 20\sqrt {10}$$