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

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Answer 1 · 2018-06-11T09:27:16

We get #a=b=4#

We have the Perimeter as

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

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

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

Answer 2 · 2018-06-11T09:44:05

#P=16#

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

#b=16/x#

measured also in inches.

The perimeter is then:

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

Evaluate the derivative:

#(dP)/dx = 2-32/x^2#

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

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

thus #x=4# is a minimum.

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