Four feet of wire is to be used to form a square and a circle

How much wire should be used for the square and how much for the circle in order to enclose the least, respectively the greatest, total area?

Your statement of the problem seems somewhat vague or unclear. However, if the problem you mean to ask is
"How can four feet of wire be used to make a square and a cricle such that the sum of their areas is (a) maximized and (b) minimized?"
then I can answer it.

Let the side length of the square be x and the radius of the circle be r. We see that the perimeter of the square is 4x and the circumference of the circle is 2(pi)r. Thus, since the sum of the perimeter and the circumference must be 4 feet, we have the equation
4x+2(pi)r=4
Let's use this equation to solve for r in terms of x:
2(pi)r=4-4x
r=(4-4x)/(2(pi))=(2-2x)/(pi)

Next, we see that the area of the square is x^2 and the area of the circle is (pi)r^2. This gives us an equation for the total area:
A=x^2+(pi)r^2
Substitute (2-2x)/(pi) for r:
A=x^2+(pi)((2-2x)/(pi))^2
A=x^2+(pi)(4-8x+4x^2)/(pi)^2
A=x^2+(4-8x+4x^2)/(pi)
A=(1/(pi))*((pi)x^2+4x^2-8x+4)
A=(1/(pi))*((4+(pi))x^2-8x+4)

This gives the area as a function of x, so we can take its derivative with respect to x and set it equal to 0 to find critical points:
dA/dx=(1/(pi))*((4+(pi))*2x-8+0)=0
(4+(pi))*2x-8=0
(4+(pi))*2x=8
x=4/(4+(pi))
This is a critical point where A(x) could have a minimum or a maximum. The other critical points are the endpoints, x=0 (all the wire goes to the circle) and x=1 (all of it goes to the square, giving it 1 ft of length per side). We must plug in these values to A(x) to see which ones result in the greatest and least total areas:
A(0)=(1/(pi))*((4+(pi))(0)-8(0)+4)
=4/(pi)
This is about 1.273.
A(1)=(1/(pi))*((4+(pi))(1)-8(1)+4)
=(1/(pi))*(4+(pi)-4)=(pi)/(pi)=1
Lastly,
A(4/(4+(pi)))
=(1/(pi))*( (4+(pi)) (4/(4+(pi)))^2-8(4/(4+(pi)))+4)