Answer:
Explanation:
Total length of the wire is 29 m.
Let the length of one piece is d and of another piece is 29 - d.
Let d is used to make a square.
And 29 - d is used to make an equilateral triangle.
(a)
Area of square = d²
Area of equilateral triangle = √3(29 - d)²/4
Total area,
[tex]A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}[/tex]
Differentiate both sides with respect to d.
[tex]\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)[/tex]
For maxima and minima, dA/dt = 0
d = 8.76 m
Differentiate again we get the
[tex]\frac{d^{2}A}{dt^{2}}= + ve[/tex]
(a) So, the area is maximum when the side of square is 29 m
(b) so, the area is minimum when the side of square is 8.76 m
