Step-by-step explanation:
Here, the total length of fencing available = 390 yd
Let L = length of the side parallel to river
W = width of other 3 sides.
So, total fencing L + 3 W = 390 yd
or, L = 390 - 3 W
Now, Area of the field = L x W
= (390 - 3 w) (W)
or, A = -3 W² + 330 W
The maximum value of above function is at W = [tex](\frac{-b}{2a} ) = \frac{-330}{2\times (-3)} = 55[/tex]
So, W = 55 yards
Now, L = (390 - 3 (55) ) = 165 yards
Now, maximized area = L x W
= 55 x 165 = 9075 sq yds