A wire 7 meters long is cut into two pieces? help soon plz!

hello

i have this calculus problem that i am getting stuck at the part with arithmatic. this is the problem:

A wire 7 meters long is cut into two pieces. One piece is bent into a equilateral triangle for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures? Give the length of wire used for each:

For the equilateral triangle:

For the circle:

Where should the wire be cut to maximize the total area? Again, give the length of wire used for each:
For the equilateral triangle:

For the equilateral triangle:

For the circle:

i am confused becaused i did what my teacher told me and i still am getting the wrong answer. its due in half an hour so any reply soon with great help will be appreciated and rewarded!

thank you!

Let x be the length of the portion used for the circle.

Then (7 - x) is the length used for the triangle.

Then the circumference of the circle is 2pi*r = x, and so

r = x/(2*pi), giving an area of pi*r^2 = x^2 / (4*pi).

Each side of the triangle will have length (1/3)*(7 - x),

so the height of the triangle will be

(1/3)*(7 - x)*sin(pi/3) = (sqrt(3) /6)*(7 - x)

and so its area will be (1/2)*(base)*(height) =

(1/2)*(1/3)*(7 - x)*(sqrt(3) /6)*(7 - x) = (sqrt(3) /36)*(7 - x)^2.

The combined area will then be

A(x) = (1/(4*pi))*x^2 + (sqrt(3) /36)*(7 - x)^2.

Now any critical points will occur when dA/dx = 0:

dA/dx = (1/(2*pi))*x - (sqrt(3) /18)*(7 - x) = 0 when

(1/(2*pi))*x = (sqrt(3) /18)*(7 - x) ------>

x*[(1/(2*pi)) + (1/18)*sqrt(3)] = (7/18)*sqrt(3) ----->

x = (7/18)*sqrt(3) / [(1/(2*pi)) + (1/18)*sqrt(3)],

which gives us x = 2.63754 m to 5 decimal places.

This is the length of the wire used for the circle,

so the length used for the triangle is 7 - x = 4.36246 m to 5 d.p..

This will yield a minimum combined area, since the

maximum area will result when all 7 m is used for the circle.

If we had made x the triangle's piece, then the combined area function
would have been A(x) = (1/(4*pi))*(x - 7)^2 + (sqrt(3) /36)*x^2.
In this case we would find that dA/dx = 0 when x = 4.36246 m.
So either way we would get the triangle's piece being 4.36246 m
and the circle's piece 2.63754 m.

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then there's 2, it takes 2 baby, you and me baby.

Please show your work.