Derivatives of Trig Functions: Applications
5. A V-shaped trough is to be constructed with sides that are 200cm long and 30 cm wide. Find the angle between the sides that maximizes the capacity of the trough.
Start with a 3D sketch of the trough and a 2D sketch of the front triangle area of the trough.
Now, what type of problem is this?
We want to mazimize capacity. What variable will we use?
Volume. What is the equation for the volume of this trough?
V = area of triangle * 200 = 1/2 bh * 200
Now look at the sketch and find expressions for b and h in terms of the angle, r.
V = 1/2 (30 sin r/2) (30 cos r/2) *200 = 90000 sin r/2 cos r/2
What is the next step?
Take the derivative and set it equal to zero.
dV/dr = 90000 sin r/2 (-sin r/2) + 90000 cos r/2 (cos r/2)
0 = 90000(-sin2 r/2 + cos2 r/2)
0 = (-sin2 r/2 + cos2 r/2)
sin2 r/2 = cos2 r/2
tan2 r/2 = 1
tan r/2 = 1
r = 45o