Answer:
a) the length of the wire for the circle = [tex](\frac{60\pi }{\pi+4}) in[/tex]
b)the length of the wire for the square = [tex](\frac{240}{\pi+4}) in[/tex]
c) the smallest possible area = 126.02 in² into two decimal places
Step-by-step explanation:
If one piece of wire for the square is y; and another piece of wire for circle is (60-y).
Then; we can say; let the side of the square be b
so 4(b)=y
b=[tex]\frac{y}{4}[/tex]
Area of the square which is L² can now be said to be;
[tex]A_S=(\frac{y}{4})^2 = \frac{y^2}{16}[/tex]
On the otherhand; let the radius (r) of the circle be;
2πr = 60-y
[tex]r = \frac{60-y}{2\pi }[/tex]
Area of the circle which is πr² can now be;
[tex]A_C= \pi (\frac{60-y}{2\pi } )^2[/tex]
[tex]=( \frac{60-y}{4\pi } )^2[/tex]
Total Area (A);
A = [tex]A_S+A_C[/tex]
= [tex]\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2[/tex]
For the smallest possible area; [tex]\frac{dA}{dy}=0[/tex]
∴ [tex]\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0[/tex]
If we divide through with (2) and each entity move to the opposite side; we have:
[tex]\frac{y}{18}=\frac{(60-y)}{2\pi}[/tex]
By cross multiplying; we have:
2πy = 480 - 8y
collect like terms
(2π + 8) y = 480
which can be reduced to (π + 4)y = 240 by dividing through with 2
[tex]y= \frac{240}{\pi+4}[/tex]
∴ since [tex]y= \frac{240}{\pi+4}[/tex], we can determine for the length of the circle ;
60-y can now be;
= [tex]60-\frac{240}{\pi+4}[/tex]
= [tex]\frac{(\pi+4)*60-240}{\pi+40}[/tex]
= [tex]\frac{60\pi+240-240}{\pi+4}[/tex]
= [tex](\frac{60\pi}{\pi+4})in[/tex]
also, the length of wire for the square (y) ; [tex]y= (\frac{240}{\pi+4})in[/tex]
The smallest possible area (A) = [tex]\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})[/tex]
= 126.0223095 in²
≅ 126.02 in² ( to two decimal places)
