(A) Minimize the perimeter of rectangles with area 25 cm^2. (B) Is there a maximum perimeter of rectangles with area 25 cm^2?

Expert Answers
jeew-m eNotes educator| Certified Educator


Let the leg lenths of rectangle be a,b.

Then area A= a*b = 25

Perimeter S = 2(a+b)


S= 2(a+b) = 2(a+25/a)

For perimeter to be max or min dS/da = 0

dS/da = 2-25/(a^2)

At max/min S dS/da = 0

2-25/(a^2) = 0

               2= 25/(a^2)

              a = +5/sqrt2  or a= -5/sqrt2


If S is a minimum then d^2S/da^2 >0

If S is a maximum then d^2S/da^2 <0

d^2S/da^2  = -25*(-2a)/a^4 = 50/a^3


If a=-5/sqrt2 (put a=-5) then [d^2S/da^2]<0

If a= 5/sqrt2 (put a=-5) then [d^2S/da^2]>0


So perimeter is minimum when a = 5/sqrt2

 Minimum S = 21.213cm



Mathematically when a= -5/sqrt2 we will get the maximum perimeter. But the actual scenario is a>0 since it is a length. So there is no maximum perimeter.