What is the smallest parameter possible for a rectangle whose area is 16 square inches and what are it’s dimensions?

2 Answers
Jun 11, 2018

We get a=b=4a=b=4

Explanation:

We have the Perimeter as

p=2(a+b)=2(a+16/a)p=2(a+b)=2(a+16a)
So we get by AM-GMAMGM

a+16/a>=2sqrt(a*16/a)=8a+16a2a16a=8
Multiplying by 22

2(a+16/a)>=4sqrt(16)=162(a+16a)416=16
the equal sign holds if
a=b=4a=b=4

Jun 11, 2018

P=16P=16

Explanation:

Let a=xa=x inches be one side of the rectangle, then the other side is:

b=16/xb=16x

measured also in inches.

The perimeter is then:

P= 2(x+16/x)P=2(x+16x)

Evaluate the derivative:

(dP)/dx = 2-32/x^2dPdx=232x2

thus the derivative is null for x=4x=4, and in this point the second derivative:

(d^2P)/dx^2 = 64/x^3 > 0d2Pdx2=64x3>0

thus x=4x=4 is a minimum.

We can conclude that the smallest perimeter is obtained when a=b=4a=b=4, that is when the rectangle is a square.