Question

how to find the minimum area

Answer 1

of anything in particular?

Answer 2

a rectangle

Answer 3

still rather vague...

essentially for any given perimeter, you can maximize area by making the rectangle more square, and minimize the area by making one side of the rectangle as long as possible and the other side as short as possible, thus making the area approach zero. However for all practical applications, that wouldnt really make much sense..

It's generally more common to maximize area for a given perimeter, though i suppose if one of the sides of the rectangle was restricted to a minimum/maximum value then you could find a minimum as well...

do you have a particular problem in mind?

Answer 4

Yes,

Find the length and width of a rectangle that has the given area and a minimum perimeter.
Given: Perimeter 80 meters

Answer 5

thas the problem

Answer 6

wait so does it have a given area or given perimeter? and minimum area or perimeter?

Answer 7

just a given perimeter
80 meters

Answer 8

alright... its worded a little bizarrely..

anyways,

if this isn't a calculus course, just realize that area will be zero if either side is 0. Thats the minimum possible area.. 0 if length or width is 40 and the opposite is 0.. Which is why it seems like a weird question. It's slightly more complicated with a maximum area, but once again, a nice noncalculus solution is that the more square a rectangle gets, the more you maximize its area to perimeter ratio. the square with perimeter 80 has sides of 20 so the area would be 20*20=4000

if this is a calculus course, you want to find critical points to find maxs and mins:
A=L*w right? and P=2L+2w
perimeter is constant so P=2L+2w=80 so L+w=40
now you can rewrite either variable in terms of the other like so:
w=40-L

Now rewrite your area equation in terms of the single variable:
A=L*w=L*(40-L)=40L-L^2
then differentiate and set it equal to 0 to find critical points:
dA/dL = 40-2L = 0

L=20.. So L=20 is your only critical point. and the only possibilities are critical points and boundaries... so either L=0 or L=20 or L=40

at L=0 or L=40, the area = 0 (minimum possible area)
at L=20, area = 4000 (maximum possible area)