MATH SOLVE

4 months ago

Q:
# You have a wire that is 44 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?

Accepted Solution

A:

Answer:The circumference of the circle is 29.36 cmStep-by-step explanation:Letx -----> the length of first piece (shape of square)y ------> the length of the other piece (shape of a circle)we know that[tex]x+y=44[/tex][tex]y=44-x[/tex] -----> equation Astep 1Find out the area of squareThe perimeter of square is equal to the length of the first piece[tex]P=4b[/tex]whereb is the length side of square[tex]P=x\ cm[/tex] [tex]x=4b[/tex][tex]b=x/4[/tex]Find the total area AThe area of square is [tex]A_1=b^2[/tex][tex]A_1=\frac{x^{2}}{16}[/tex]step 2Find out the area of the circleThe circumference of the circle is equal to the length of the other piece[tex]C=(44-x)\ cm[/tex]The circumference is equal to[tex]C=2\pi r[/tex]so[tex]2\pi r=(44-x)\ cm[/tex]Find the radius of the circle[tex]r=\frac{(44-x)}{2\pi}\ cm[/tex]Find the area of the circle[tex]A_2=\pi r^{2}[/tex]substitute the value of r[tex]A_2=\pi (\frac{(44-x)}{2\pi})^{2}[/tex][tex]A_2= \frac{(44-x)^2}{4\pi}[/tex]step 3Find out the total area[tex]A=A_1+A_2[/tex]substitute[tex]A=\frac{x^{2}}{16}+\frac{(44-x)^2}{4\pi}[/tex]This is a vertical parabola open upwardThe vertex is a minimumusing a graphing toolThe vertex is the point (14.64, 67.77)see the attached figureFor x=14.64 cm -----> the area is a minimumThe lengths of the wire are[tex]x=14.64\ cm\\y=44-14.64=29.36\ cm[/tex]thereforeThe circumference of the circle is 29.36 cm