Hi, I'm having doubt about whether or not I'm approaching two related rates of change problems correctly... Below is my working: 1) The diagram represents a trough. (It's drawn like a cylinder, except the ends are right, isosceles triangles) The trough is 1.5m long. Water is being poured into the trough at the rate of 4litres per minute. At what rate is the water in the trough rising after 5 minutes? Now, here's how I approached this one: I first derived the volume formula from the information given. The formula would be area of triangle times by 150cm (Converting the length) and because it's isosceles (the triangle) the height is equal to the width, so the volume V = 150 * 0.5 * xy = 150 * 0.5 * x^2 = 75x^2 That's out of the way! Now, it says water is being poured into the trough at 4L per minute, which leads me to interpreting it as: dV/dt = 4L/min = 4000cm^3 /min (converting to centimetres cubed) Now, it asks for the rate the water's rising after 5 minutes, dx/dt (Again, my understanding), so we'll find the volume after 5 minutes in order to get a proper value for the height to find dx/dt: dV/dt * 5 = 20000cm^3 <-- The volume after 5 minutes.. And I have a volume formula, so let's make it equal to the volume after 5 minutes: 20000cm^3 = 75x^2 Solving for x I get x, height & width, = 16.3299cm (To 4dp) Now's the question of finding dx/dt... I already have dV/dt, which is 4000cm^3/min, and I have dV/dx from V, which is 150x, so let's use the chain rule and some algebra: dV/dt = dV/dx * dx/dt 4000 = 150x * dx/dt Now, I know what the height will be after 5 minutes, 16.3299cm, so I have... dx/dt = 4000/150x = 4000/150(16.3299) = 1.63299cm/min <-- The rate of rising of the water in the trough after 5 minutes. ....or so I think; did I screw up? :\ 2) A rectangular carton is to be constructed so that the length is twice the breadth. The total surface is to be 1200 square centimetres. Find the dimensions of the box that will give the greatest volume. I derived a surface area equation using the geometric properties of a cuboid, ending up with the formula: SA = 4x^2 + 6xy (Where x = breadth, y = height) And given the fact that the surface area will be 1200cm^2, I made it equal: 4x^2 + 6xy = 1200 Now, what I needed was y in terms of x so I can get myself a single-variable volume formula; using algebra: y = (600-2x^2)/3x Now, thinking about it, volume is the area of the base, 2x * x, multiplied by the height, y, so: V, volume, = 2x^2y Substituting for y, after some simplification and muliplying out of brackets, I get: V = 400x - 4/3 x^3 Now, we want maximum dimensions for maximum area, so differentiating the volume formula: dV/dt = 400 - 4x^2 => 0 4x^2 = 400 x^2 = 100 x = 10cm There, we have one dimension that would give the greatest volume! Now for another, 2x: 2x = 20cm Now for the third, y: y = 400/30 = 13.333333cm (No clue how to put the recurring sign in LaTeX) So, I'm left with the maximum dimensions: 10cm x 20cm x 13.33333333cm That's what I think the answer is. I don't know if what I've come up with is right or wrong, but I'd like someone to point out if I made a huge mistake... Can anybody please shed some light on if I've done everything correct or not? And if I have made errors, please tell me where I screwed up. Thanks a lot for any attempts at helping me out; I appreciate it.