Answer :
Answer:
I = 60°
Step-by-step explanation:
Step 1: Bending one third of the sheet
[tex]\frac{1}{3} * (30) = 10 cm[/tex]
So there are two segments of 10 cm on both ends that are bent upwards and a flat part of 10 cm in the center.
The more the area of the sheet the more the flow of the water. We need to find a relation between the area of the sheet and angle.
Step 2: Area of the sheet after bending can be written as
[tex]A = 10h + 2(\frac{1}{2} bh )[/tex]
Where h is the height of the bent up segments above the base
Step 3: Relation between Area and angle I
sin I = [tex]\frac{h}{10}[/tex] ; h = 10 sin(I)
cos I = [tex]\frac{b}{10}[/tex] ; b = 10 cos(I)
Therefore Area in terms of angle I can be written as
[tex]A(I) = 10(10 sin(I)) + 2(\frac{1}{2} (10 sin(I)) (10 cos(I)))[/tex]
[tex]A(I) = 100 sin(I) + 100 sin (I) cos (I)[/tex]
Step 4: Finding the maximum area.
To find the maximum are we need to differentiate the above equation and find the critical points of the equation
[tex]A'(I) = 100 cos(I) + 100[cos(I)cos(I)+ (-sin(I)sin(I))][/tex]
[tex]A'(I) = 100 cos(I) + 100[cos^{2}(I) - sin^{2} (I)][/tex]
[tex]A'(I) = 100[ cos(I) + cos^{2}(I) - sin^{2} (I)][/tex]
using trigonometric relation
[tex]sin^2 (x) + cos^2 (x) = 1[/tex]
[tex]sin^2 (x) = 1 - cos^2 (x)[/tex]
therefore equation A'(I) becomes
[tex]A'(I) = 100[ cos(I) + cos^{2}(I) - (1 -cos^{2}(I) )][/tex]
[tex]A'(I) = 100[ cos(I) + cos^{2}(I) - 1 +cos^{2}(I)][/tex]
[tex]A'(I) = 100[2cos^{2}(I)+cos(I) -1][/tex]
Factorizing the above equation gives
[tex]A'(I) = 100[2cos^{2}(I)-cos(I) + 2cos(I) -1][/tex]
[tex]A'(I) = 100[cos(I)(2cos(I)-1) + 2cos(I) -1][/tex]
[tex]A'(I) = 100[(cos(I)+1) (2cos(I)-1)][/tex]
To find the critical points we need to equate the above the equation to zero
[tex]100[(cos(I)+1) (2cos(I)-1)] = 0[/tex]
Therefore,
[tex]cos(I)+1 = 0[/tex] [tex]or[/tex] [tex]2cos(I)-1 = 0[/tex]
I = 60° or 180°
Angle I can be either 60 or 180 for which the area is maximum, therefore we have to calculate the area using both of these angles to see which of them gives us the desired result.
Using [tex]A(I) = 100 sin (I) + 100 sin (I) cos (I)[/tex]
When I = 60°
[tex]A(60) = 100 sin (60) + 100 sin (60) cos (60)[/tex]
[tex]A(60) =129.9 cm^2[/tex]
When I = 180°
[tex]A(180) = 100 sin (180) + 100 sin (180) cos (180)[/tex]
[tex]A(180) = 0[/tex]
Therefore, the angle for which the sheet can carry the maximum water is 60° and the resulting area is 129.9 cm².