Answer:
a) the length of the wire for the circle =
b)the length of the wire for the square =
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=
Area of the square which is L² can now be said to be;
On the otherhand; let the radius (r) of the circle be;
2πr = 60-y
Area of the circle which is πr² can now be;
Total Area (A);
A =
=
For the smallest possible area;
∴
If we divide through with (2) and each entity move to the opposite side; we have:
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
∴ since , we can determine for the length of the circle ;
60-y can now be;
=
=
=
=
also, the length of wire for the square (y) ;
The smallest possible area (A) =
= 126.0223095 in²
≅ 126.02 in² ( to two decimal places)
