Find the minimum premeter for a rectangle?

[Brown Arrow]

Homework Statement

A rectangular field to contain a given area is fenced off, along a stright river. if no fencing is required along the river, show that the lest amount of fencing will be required when the length of the field is twice its width.

Homework Equations

N/A

The Attempt at a Solution

i really don't get how to solve this,
my attempt:

Please help

 

[SammyS]

You want to maximize the Area, not the Perimeter !

 

[Brown Arrow]

but they said they want to minimize the fencing.

even if i did A=lXW, i would end up with A=2w²
if i derive it i get A'=4W, and W=0; ?? i don't get it sorry.

 

[SVXX]

@Sammy : The area is given, isn't it?? Why would he want to maximize the area? He has to use the given relation for the area, put it in the perimeter relation and then differentiate it with respect to either length or breadth...find stationary points(where d/dl or d/db is 0) and then use a double differential test to check for maxima or minima.
Also, note how one side will not be fenced. That means a little change in the perimeter relation.

That means something like P = L + 2B and A = L * B. Put A in P to get P = L + 2A/L where A is constant. Find dP/dL and proceed as I said.

 

[Brown Arrow]

so if do it like this:

but i still don't get it, are we not using the fact l=2w, if we did here i would get w=(A_o/(2))^1/2

then it doesn't make any sense

 

[SammyS]

@Sammy : The area is given, isn't it?? Why would he want to maximize the area? He has to use the given relation for the area, put it in the perimeter relation and then differentiate it with respect to either length or breadth...find stationary points(where d/dl or d/db is 0) and then use a double differential test to check for maxima or minima.
Also, note how one side will not be fenced. That means a little change in the perimeter relation.

That means something like P = L + 2B and A = L * B. Put A in P to get P = L + 2A/L where A is constant. Find dP/dL and proceed as I said.

Yes, I suppose that's correct.

 

[SammyS]

Then A is constant, so A' = 0.

Then P' = 2-(A/W²).

but P' = 0 at min.

0 = 2-(A/W²).

Solve for W.

This does work out!