A wire 10m long is cut into 2 pieces. one piece will be cut into a square the other piece will be shaped into a circle. Graph A= A(x) for what value of x is A smallest?

Answer 1

if the side of the square is x,
square has perimeter 4x
circle has circumference 10-4x

A(x) = x^2 + pi * ((10-4x)/(2pi))^2
= 1/pi ((4+pi)x^2 - 20x + 25)

dA/dx = 2/pi ((4+pi)x - 10)
dA/dx = 0 when x = 10/(4+pi)
since the parabola opens up, A is smallest there.

Answer 2

i don't understand how to graph what is x and y

Answer 3

As I said, let x be the side of the square. The perimeter is thus 4x.

The available length of wire is 10. So, the wire available for the circle is only 10-4x

area of square is x^2
area of circle is pi r^2
if the circumference of the circle is (10-4x), the radius is (10-4x)/2pi

so, the area of the combined figures is as shown above:

A(x) = 1/pi ((4+pi)x^2 - 20x + 25)
now, this is pre-cal, so you know about parabolas. You know that the vertex of a parabola ax^2+bc+c is at x = -b/2a.

Here, that would be x = 20/(2(4+pi)) = 10/(4+pi)

Since the coefficient of x^2 is positive, the parabola opens up, so the value of A(x) at the vertex is a minimum.