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)

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)

$\displaystyle P$ is a constant ...

$\displaystyle P = 2L + 2W$

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

$\displaystyle A = LW$

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

note that $\displaystyle A$ is quadratic in $\displaystyle L$ ... you should find the max area occurs when the rectangle is a square.

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.

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.

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

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?

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).

Re: intermediate and extreme value theorms?

would the first or second derivative test for extrema suffice?

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?

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