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

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### 1 Answer

A)

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**

B)

**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.**