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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?

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    $\begingroup$ Solving for the critical point $x^2=9000$ does not give $x=30$, you've lost one zero in $9000$. $\endgroup$ – A.Γ. Jul 31 at 22:40
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You made a computational error. Actually, $x=30\sqrt{10}$ and $y=20\sqrt{10}$.

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    $\begingroup$ I believe you have your $x$ and $y$ values mixed around. $\endgroup$ – John Omielan Jul 31 at 22:40
  • $\begingroup$ Yes. I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Jul 31 at 22:51
  • $\begingroup$ Does the dividing fence factor in at all to my calculations? Why don't I just find the absolute minimum like usual - ignoring the extra fence? I can add that in after. $\endgroup$ – Burt Jul 31 at 23:01
  • $\begingroup$ If you forget the dividing fence factor, you will get a different answer (namey, a square), and clearly both solutions can't be right. $\endgroup$ – José Carlos Santos Jul 31 at 23:03
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    $\begingroup$ It is so because you are changing the problem and therefore it is natural that you get a different answer. And it is a minimal perimeter, since the second derivative at $30\sqrt{10}$ is positive. $\endgroup$ – José Carlos Santos Jul 31 at 23:18
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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 $$y= 20\sqrt {10}$$

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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 $P$.

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    $\begingroup$ Welcome to MathSE. To write $\sqrt{4000}$, type $\sqrt{4000}$. This tutorial explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Jul 31 at 22:53
  • $\begingroup$ Thank you for the tip! $\endgroup$ – Amy Ngo Jul 31 at 22:56

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