Answer:
Rectangular path
Solution:
As per the question:
Length, a = 4 km
Height, h = 2 km
In order to minimize the cost let us denote the side of the square bottom be 'a'
Thus the area of the bottom of the square, A = [tex]a^{2}[/tex]
Let the height of the bin be 'h'
Therefore the total area, [tex]A_{t} = 4ah[/tex]
The cost is:
C = 2sh
Volume of the box, V = [tex]a^{2}h = 4^{2}\times 2 = 128[/tex] (1)
Total cost, [tex]C_{t} = 2a^{2} + 2ah[/tex] (2)
From eqn (1):
[tex]h = \frac{128}{a^{2}}[/tex]
Using the above value in eqn (1):
[tex]C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}[/tex]
[tex]C(a) = 2a^{2} + \frac{256}{a}[/tex]
Differentiating the above eqn w.r.t 'a':
[tex]C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}[/tex]
For the required solution equating the above eqn to zero:
[tex]\frac{4a^{3} - 256}{a^{2}} = 0[/tex]
[tex]\frac{4a^{3} - 256}{a^{2}} = 0[/tex]
a = 4
Also
[tex]h = \frac{128}{4^{2}} = 8[/tex]
The path in order to minimize the cost must be a rectangle.