A rectangle has a surface area of 100 m². The perimeter of it can be written as the following.

Answer 1

LW = 100

(a)

W = 100 / L

P = 2(W + L)

P = (200 / L ) + 2L

(b)

L = 100 / W

P = 2(W + L)

P = 2W + (200 / W)

Answer 2

p=(2s²+200)/s,,,where s is one side of rectangle.

p=2s₁+2s₂,,,,,s₂=100/s₁

to check. know 20 by 5 rectangle has area = 100

perimeter is 50=40+10

let s= 5 by formula, p=(2*25+200)/5 =50

let s=20 by formula p=(2*400+200)/20 =50

if calculus is used

dp(s)/ds=2-200s^-2,,,,,,,,,,,,is first derivative

dp(s)/ds=0 is critical point

dp(s)/ds=0 when s=10,,,,,,,,,,,-10 is not physically possible

the suspicion is s=10 gives minimum perimeter since s=10 means p=40

d²p(s)/ds²=400s^-3 is second derivative

and second derivative is positive at s=10,

so s=10 is side length for minimum perimeter