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A fun little question which I'm afraid has got me stumped. I'm not really sure how to approach this, what do you think?
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hi Au101,
I think the formula would be something like:
The given values should be enough to calculate the constants j, k and L.
Then you can differentiate to get the turning points, and show the one you find is a minimum, not a maximum.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks a lot! That seems to've done the trick
I was about halfway there, but I hadn't quite fully understood the question.
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Hopefully, that's what the questioner had in mind. The graph cost against V will have a minimum so you should be able to solve this now.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Yeah - well, certainly, the answer checks out (26 when rounded appropriately).
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The final question of this exercise is (in my opinion) another slightly fiendish one. Again, there's something I'm just not getting. Here it is:
For what it's worth, I've worked out that the gradient (call it m) of the line is:
But I can't say I've had any more breakthroughs than that.
I wondered briefly whether the least value of OA + OB will occur when the gradient is -1. If that's true I haven't been able to prove it, or get anywhere by taking that as a starting point!
Last edited by Au101 (2015-02-10 11:08:30)
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The problem is now to treat m as a variable find the value of it that minimizes OA+OB.
Similarly for the triangle, find the value of m that minimizes (OA·OB)/2.
Last edited by Olinguito (2015-02-10 21:29:39)
Bassaricyon neblina
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Thank you very much Olinguito, that's got it, I just needed help with how to think about it
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