Thread: 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.

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.

is a constant ...









note that is quadratic in ... you should find the max area occurs when the rectangle is a square.

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.

it's a optimization problem ... find the value of L in terms of P that maximizes the area of the rectangle.

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

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?

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).

would the first or second derivative test for extrema suffice?

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?

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