# intermediate and extreme value theorms?

• Oct 22nd 2011, 02:09 PM
Barthayn
intermediate and extreme value theorms?
I need help with starting the question below. I believe I have to use either intermediate-value or extreme-value theorems, however, I am not sure. The question is as follows:

Fix positive number P. Let R denote the set of all rectangles with perimeter P. Prove that there is a member of R that has maximum area. What are the dimensions of the rectangle of maximum area? HINT: Express the area of an arbitrary element of R as a function of length of one of the sides.

No idea where to begin. I understand it wants the max area and the sides of it. However, I don't know what to do. (Headbang)
• Oct 22nd 2011, 02:42 PM
skeeter
Re: intermediate and extreme value theorms?
Quote:

Originally Posted by Barthayn
I need help with starting the question below. I believe I have to use either intermediate-value or extreme-value theorems, however, I am not sure. The question is as follows:

Fix positive number P. Let R denote the set of all rectangles with perimeter P. Prove that there is a member of R that has maximum area. What are the dimensions of the rectangle of maximum area? HINT: Express the area of an arbitrary element of R as a function of length of one of the sides.

No idea where to begin. I understand it wants the max area and the sides of it. However, I don't know what to do. (Headbang)

$P$ is a constant ...

$P = 2L + 2W$

$W = \frac{P-2L}{2}$

$A = LW$

$A = L\left(\frac{P-2L}{2}\right) =$

note that $A$ is quadratic in $L$ ... you should find the max area occurs when the rectangle is a square.
• Oct 23rd 2011, 04:11 PM
Barthayn
Re: intermediate and extreme value theorms?
I still don't understand where to go from here. I know to understand where to get the lengths of the sides of the rectangle I should take A(L) = A'(L). However, I do not see how this is proof and where this will lead to.
• Oct 23rd 2011, 04:35 PM
skeeter
Re: intermediate and extreme value theorms?
it's a optimization problem ... find the value of L in terms of P that maximizes the area of the rectangle.
• Oct 23rd 2011, 04:53 PM
Barthayn
Re: intermediate and extreme value theorms?
I understand that. This is what I do:

A = LW
P = 2L+2W
A(L) = 1/2LP - L^2
A'(L) = 1/2P - 2L
A'(L) = 1/2P - 2L = 0
A'(L) => 1/4P

From here, now knowing that L = 1/4 P I do not know where to go. I get 1/4P if I sub it into P = 2L+2W and zero if I put it into A(L) = 1/2LP - L^2
• Oct 23rd 2011, 05:00 PM
skeeter
Re: intermediate and extreme value theorms?
Quote:

Originally Posted by Barthayn
I understand that. This is what I do:

A = LW
P = 2L+2W
A(L) = 1/2LP - L^2
A'(L) = 1/2P - 2L
A'(L) = 1/2P - 2L = 0
A'(L) => 1/4P

From here, now knowing that L = 1/4 P I do not know where to go. I get 1/4P if I sub it into P = 2L+2W and zero if I put it into A(L) = 1/2LP - L^2

if the length is P/4, what does that say about the lengths of the other 3 sides?
• Oct 23rd 2011, 05:07 PM
Barthayn
Re: intermediate and extreme value theorms?
it says that there are all 1/4 P. !!!! It is a perfect square!!! Therefore it has maximum area. However, how do I prove that this claim is true? (my university always wants people to prove it mathematically, otherwise you mention a proven theorem that was taught in class).
• Oct 23rd 2011, 05:41 PM
skeeter
Re: intermediate and extreme value theorms?
would the first or second derivative test for extrema suffice?
• Oct 23rd 2011, 05:53 PM
Barthayn
Re: intermediate and extreme value theorms?
not sure. I would say the first derivative because it tells you when something is at a maximum (in this case the area). How would I should it to be that case though?
• Oct 23rd 2011, 05:57 PM
skeeter
Re: intermediate and extreme value theorms?
Quote:

Originally Posted by Barthayn
not sure. I would say the first derivative because it tells you when something is at a maximum (in this case the area). How would I should it to be that case though?

show A'(L) > 0 for L < P/4 and A'(L) < 0 for L > P/4

second derivative test may be easier ... all you would have to show is that A''(P/4) < 0