Expert: Paul Klarreich - 2/6/2008

Question
A farmer from the Borders is taking as many of sheep to market as can possibly be penned in with 40 meters of flexible fencing.
1) If he uses the corner where two long straight walls meet as two of the sides of his rectangle, what is the largest area? Assume the walls meet at a right angle.
2) If he uses the corner as two of the sides and two straight lengths of fence to form a quadrilateral, rather than a rectangle, investigate the area enclosed to try to maximize it.  
3) The farmer now decides to try out making his pen rectangular: if he uses the corner for two of the sides and the 40m fence as the third side, what are the angles of the triangle for which the area is maximum. What is this area?

Answer
Questioner:   Henoc
Category:  Advanced Math
Private:  No
 
Subject:  max problems, areas...
Question:  A farmer from the Borders is taking as many of sheep to market as can possibly be penned in with 40 meters of flexible fencing.
1) If he uses the corner where two long straight walls meet as two of the sides of his rectangle, what is the largest area? Assume the walls meet at a right angle.
2) If he uses the corner as two of the sides and two straight lengths of fence to form a quadrilateral, rather than a rectangle, investigate the area enclosed to try to maximize it.  
3) The farmer now decides to try out making his pen rectangular: if he uses the corner for two of the sides and the 40m fence as the third side, what are the angles of the triangle for which the area is maximum. What is this area?
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Hi, Henoc,

Since you  didn't indicate what, if anything, you already did on these, I will assume you just don't know how to get started.

1. Let x, y be the sides.  Use A = xy  y = 40 - x.  Then maximize A, using the fact that y = 40 - x.

2. This is rather vague. Investigate?  How about assuming that the two lengths will be equal and just varying their angle -- the one opposite the corner.  This will take a bit of analysis.

3. I think you mean 'making his pen "triangular"', not rectangular.  Now:

Let ABC be the triangle, with C the Corner.  Then the variable you want is t = angle BAC, and you will have  a, b, c as the sides, with:

c = 40
a = 40 sin t
b = 40 cos t

A = ab/2 = 800 sin t cos t  (which is also  400 sin 2t)

Now maximize A in terms of t.