Figure 1.
We denote the width of the pen by a and its length by b, as shown in Figure 1.
The area of the pen is ab, and we have 2a + b = 60. So, replacing b by 60 - 2a, we find that the area is a(60 - 2a), which is 60a-2a2.
We differentiate this with respect to a, getting 60 - 4a. Setting this to 0 to find the maximum (or maybe minimum) area, we find a=15, b=30.
Solution using symmetry: show
The farmer "borrows" another 60 hurdles. She will mentally build as big a pen as possible
with all 120 hurdles, situate it so that the fence runs through its centre, and then "return"
the 60 hurdles that are on the other side of the fence. She realises that the biggest pen she
can build with 120 hurdles is a square, 30 hurdles to a side. But there are many possibilities
for the orientation of the pen relative to the fence, as shown:
Figure 5.
Figure 4.
Figure 3.
Figure 2.
All of these must have a long side of 30 hurdles, with the remaining hurdles split in any way
(including 30-0 and 15-15) between the two short sides, which are necessarily at right angles to the
long side.
Comparison of solutions: show
I prefer the symmetry method to the calculus method for several reasons.
- It is easily grasped. I can do it my head. I don't need paper.
- It is less subject to errors. People sometimes make mistakes when manipulating
equations and differentiating terms.
- It is convincing. Once you have seen it you know that it is right, there
is nothing that needs checking.
- It makes it clear why half the hurdles must be used for the long side,
rather than just churning it out as a result.
- It generates all the solutions, not just one of them.
The farmer "borrows" another 60 hurdles. She will mentally build as big a pen as possible with all 120 hurdles, situate it so that the fence runs through its centre, and then "return" the 60 hurdles that are on the other side of the fence. She realises that the biggest pen she can build with 120 hurdles is a square, 30 hurdles to a side. But there are many possibilities for the orientation of the pen relative to the fence, as shown:
Figure 5.
Figure 4.
Figure 3.
Figure 2.
All of these must have a long side of 30 hurdles, with the remaining hurdles split in any way (including 30-0 and 15-15) between the two short sides, which are necessarily at right angles to the long side.
I prefer the symmetry method to the calculus method for several reasons.
- It is easily grasped. I can do it my head. I don't need paper.
- It is less subject to errors. People sometimes make mistakes when manipulating equations and differentiating terms.
- It is convincing. Once you have seen it you know that it is right, there is nothing that needs checking.
- It makes it clear why half the hurdles must be used for the long side, rather than just churning it out as a result.
- It generates all the solutions, not just one of them.