Answer:
60
Explanation:
The volume of the water carried will be proportional to the cross-sectional area. The cross-section is a trapezoid with height 10sint, where t represents theta. The top of the trapezoid is 10+(2)(10cost), i.e. 10 + 20cost. The base of the trapezoid is 10.
Area of trapezoid = (average of bases) times (height)
= ([10 + (10 +20cost)] / 2 ) 10 sint
= (10 + 10cost)(10sint)
= 100sint + 100sintcost, call this A(t). You need to maximize. So differentiate and set equal to zero.
dA/dt = -100sin(t)^2+100cos(t)+100cos(t)^2 = 0, divide by 100:
-sin(t)^2 + cos(t) + cos(t)^2 = 0, replace sin^2 by 1-cos^2
2cos(t)^2 + cos(t) - 1 = 0, factor
(1+cos(t))(2cos(t)-1)=0, so
cos(t)=1, t=0 that give a min (zero area) not a max, or
cos(t) = 1/2, so t=60 degrees. This gives the max.
