You have a wire that is 23 cm long. you wish to cut it into two pieces. one piece will be bent into the shape of a square. the other piece will be bent into the shape of a circle. let a represent the total area of the square and the circle. what is the circumference of the circle when a is a minimum?

In explanation you see the answer but it I see it send it  different  so I just added a file so you can see clearly the way how this exercise works. Please, see the file attached

Explanation:

Let x = circumference of the circle

then

(53-x) = the perimeter of the square

and \frac{53-\times}{4} = the side of the square

and

\left(\frac{53-\times}{4}\right)^2 = the area of the square

:

Find the area of the circle using the circumference

find the radius (r)

2*pi*r = x

r = \frac{\times}{2*\mathbit{\pi}}

r = \frac{\times}{6.28}

Find the area of the circle

\mathbit{A}=\pi r^2

Replace r with \frac{\times}{6.28}

A = π * \left(\frac{\times}{6.28}\right)^2=  * \left(\frac{\times^2}{39.48}\right)      

cancel pi into 39.48

A =  \left(\frac{\times^2}{12.566}\right)

Total area of circle and square

A =  \left(\frac{\times^2}{12.566}\right)+\left(\frac{53-\times}{4}\right)^2  = \left(\frac{\times^2}{12.566}\right)  + \left(\frac{{2809-106\times+\times}^2}{16}\right)

:

convert these fractions to decimal coefficients

A(x) = .0796x^2 + .0625x^2 - 6.625x + 175.5625

A(x) = .1421x^2 - 6.625x + 175.5625

:

Find the axis of symmetry of this quadratic equation (min area)

x =   \frac{-\left(-6.625\right)}{2*.1421}

x =   \frac{6.625}{.2842}

x = 23.31 cm is the circumference when they have min area

sorry? idk what i'm supposed to exain and i might

a and b because with the other three, they simplify to something else.