Let x be the length of wire allocated for the circle
Then 20-x is allocated for the square
A_circle =where --> A_circle =
A_square ==
A = A_total =+ (*)
= +
[ Note= ---> concave up --> setting dA/dx=0 will find a MINIMUM ]
= + = 0
or approx. ft
A_min =+
=+
=
For A_max, you must check endpoints (i.e. x=0ft and x=20ft). Here you will find that if you allocate all of the wire for the circle then the area will be maximized.
This graph of A_total (eq (*)) illustrates this:
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