Okay, so you don't want "complete answers", you want suggestions on how to approach the problems.

Differentiate 3V = πR^3 sin^2 θ cos θ with respect to θ and set it equal to 0. Solve for θ (the question implies that it might be easier to solve for tan θ first.) Since the derivative of V, for this θ is 0, there may be a max or min here. Show it is a max by looking at the second derivative and using the "second derivative test"."A sector S of a circle with radius R whose angle at the centre of the circle is φ radians, is rolled up to form the curved surface of a right cone standing on a circular base. The semi-vertical angle of this cone is θ radians. Express φ in terms of sin θ and show that the volume V of the cone is given by 3V = πR^3 sin^2 θ cos θ. If R is constant and θ varies, ﬁnd the positive value of tan θ for which dV/dθ = 0. Also, show further that when this value of tan θ is taken, the maximum value of V is obtained. Therefore show that the maximum value of V is( 2πR^3√3)/27 "

No, you are supposed to come up with your own formula! The volume of a cylindrical solid is the length of the cylinder times the base- and here the base is a triangle so that should be easy.2) A water trough with vertical cross-section in the form of an equilateral triangle is being filled at a rate of 4 cubic metres per minute. Given that the trough is 12 metres long, how fast is the level of the water rising when the water reaches a dept of 1.5 metres."

With no formula to work with, am I supposed to graph it first or something?

Thanks guys!

But isn't there something missing here? Is nothing said about large the equilateral triangle is? For the moment, let the length of the sides of the equilateral triangle cross section be "L".

Start by drawing a picture. The base of the triangle (here, the top) forming the trough has length L and its height is (Pythagorean theorem). The water will form a triangle sharing the bottom vertex and so is "similar" to the entire side of the trough. If the depth of the water is "h" then the ratio of height to base, h/b, for the water is the same as that ratio for the entire side of the trough: so and . Since the area of a triangle is (1/2)bh, you can write the area of a "side" of the water in terms of h only and then multiply by the length of the trough to get the volume.