A) A rectangular pen is built with one side against a barn, 200 meters of fencing are used for the other three sides of the pen. What dimensions maximize the area of the pen?

B) A rancher plans to make four identical and adjacent rectangular pens against a barn, each with an area of $100\space \text{m}^2$. What are the dimensions of each pen that minimize the amount of fence to be used?

  • $\begingroup$ Is this homework? What have you tried? $\endgroup$ – Jaycob Coleman Nov 7 '13 at 22:09
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    $\begingroup$ I drew a picture of a barn the fence with three sides I labeled two sides x and one side y, 2x+y=200. I'm a bit stuck I know I have to use application of a derivative, I'm familiar with a similar problem but this is a 3 sided fence. $\endgroup$ – ashabasha Nov 7 '13 at 22:19
  • $\begingroup$ Keep in mind that $A=xy$ still and we can express $y$ in terms of $x$. $\endgroup$ – Jaycob Coleman Nov 7 '13 at 22:23

A) Let the fence form a the $3$ sides of a rectangle; with the side of the barn being the $4^{th}$ side. Also let $x$ be the width of the rectangle and $y$ be its length. Clearly $2x+y=200 ....(1), \space \text{and the area enclosed (A) is given by:}\space A=xy....(2)$

Replacing $y$ in (2) by the expression in (1): $$\begin{align} A=xy=x(200-2x)= &200x-2x^2 \\ \frac{dA}{dx}=200-4x \\ \end{align}$$ To maximize the area, $\frac{dA}{dx}=0$ which is equivalent to $x=50 ; y=100$

B) Same situation here, except $A=4xy=400$ $$\begin{align} L=\text{total length of fence needed}&=5x+4y \\ \\ &= 5x+\frac{400}{x} \\ \frac{dL}{dx}=5-\frac{400}{x^2}\\ \end{align}$$ To minimize L, $\frac{dL}{dx}=0; \space \text{which is equivalent to} \space x=\sqrt{80}=4\sqrt{5} \approx 8.94; \text{and} \space y=5\sqrt{5} \approx 11.18 $


The question is not uniquely solvable because we don't know the barn's dimensions.

  • $\begingroup$ That's not true. @K.Rmth gives the correct solution. The trick is putting your $y$ lengths in terms of the given perimeter and $x$, then finding critical points. $\endgroup$ – Jaycob Coleman Nov 7 '13 at 22:37
  • $\begingroup$ That is, under the assumption that the barn is as large as necessary for the maximized area, which is typically assumed for these problems. That this should be part of the question statement is certainly debatable. $\endgroup$ – Jaycob Coleman Nov 7 '13 at 22:52
  • $\begingroup$ So the farmer's barn in 100 meters long, maybe even longer? Lucky him! Btw: from $A=x(200-2x)$ you conclude immediately $x=50$. That's the “trick.” No calculus needed. $\endgroup$ – Michael Hoppe Nov 8 '13 at 6:11
  • $\begingroup$ Indeed it isn't necessary, but if you take a look at the other question OP has asked and the fact that both are tagged as calculus and ask essentially about finding critical points, it's pretty obvious that OP is a calc I student looking for help with how to solve basic maximization/minimization problems using calculus. $\endgroup$ – Jaycob Coleman Nov 8 '13 at 7:32

Derivatives are not needed for this problem.

The length restriction:$$2x+y=200$$The area restriction:$$A=x\cdot y$$Combining:$$2x+\frac{A}{x}=200$$ $$2x^2-200x+A=0$$The discriminant of this quadratic is:$$D=(-200)^2-8A$$Now, what is the maximum value of A that still results in a real solution?

The same procedure will solve Part B, with the appropriate area and length expressions, and considering the minimum $L$ that permits a real solution.

Edit to address comment;

For part B

The length restriction:$$5x+4y=L$$The area restriction:$$100=x\cdot y$$Combining (and multiplying through by $x$):$$5x+\frac{4\cdot 100}{x}=L$$ $$5x^2-Lx+400=0$$The discriminant of this quadratic is:$$D=(L)^2-4\cdot 5\cdot 400$$Now, what is the minimum value of L that still results in a real solution?

  • $\begingroup$ It's not the same procedure. $\endgroup$ – Michael Hoppe Nov 8 '13 at 7:35
  • $\begingroup$ @MichaelHoppe: Looks very similar... I did a lot of cut-and-paste anyway... $\endgroup$ – DJohnM Nov 8 '13 at 19:32

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