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    Optimization problem - right circular cylinder inscribed in cone

    1. You know the expression for the volume of the cylinder in terms of r and h - that's dead easy. The problem is it has two variables, h and r so is tricky to differentiate - it needs partial differentiation with respect to both variables which you have probably not yet done. So make it...
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    Optimization problem - right circular cylinder inscribed in cone

    Another point. If you maximise the volume of the cylinder you just have the one formula to work with - the volume of the cylinder. And it is very easy to see what the variables in the problem are. If you minimise the volume left in the cone you have to work with the volume of the cylinder as...
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    Optimization problem - right circular cylinder inscribed in cone

    Either method is fine and making the volume of the "remaining cone minus cylinder" a minimum will work. That logic is fine. It's when you start calculating you go wrong. You are jumping ahead too fast and making mistakes. You need to take simple steps where you know each step is correct...
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    Prove .999 repeating = 1

    That is why the proof starts by "divide 3 into 1" instead of saying "consider 0.33333 ...". Any person who can divide by 3 should be able see that 1 divided by 3 is exactly equal 0.333333..., going on to infinity (my ... says goes to infinity), and that the only digit which can appear is 3. If...
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    Prove .999 repeating = 1

    The simple "high school proof" takes only two lines. divide 3 into 1 to get 1/3 = 0.333333... multiply both sides of the equation by 3 to get 1 = 0.999999... or divide 9 into 1 to get 1/9 = 0.111111... multiply both sides of the equation by 9 to get 1 = 0.9999999...
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