来自《耶鲁大学开放课程：博弈论 第14集 第14集》

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  this isn't the only way to do it--I'm sorry 150
  这不是唯一的方法 抱歉 是150
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  --this isn't the only way that we could do it,
  这不是解决这个问题的唯一方法
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  we could also set up a Lagrangian equation,
  你同样可以通过建立拉格朗日公式解答
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  but for those people who don't know what that is,
  但是对于那些不知道这个的同学
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  don't worry we're going to plug in today.
  不要担心 我们只要做代入运算就行了
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  So we're going to plug this in, and when we plug it in,
  所以现在我们把它代入 之后你会发现
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  we get a right old mess but let's do it anyway.
  我们又看到了老一套 继续
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  So we're going to get max Q1, [A - B Q1--B [[A - C]/2B - Q1/2] - C] Q1
  所以我们要取得最大值Q1 [A-B Q1--B [[A-C]/2B-Q1/2]-C] Q1
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  Everyone okay with that?
  所有人都明白了吗
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  I'm doing algebra on the board which is not fun
  我现在要在黑板上做无聊的代数了
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  but it's useful to do occasionally.
  但是在这道题里真的很有用
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  So eventually what are we going to do?
  所以最终我们要干嘛