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q1 So let me just rewrite that as, I just want to divide by q1 everywhere.
分子分母同时除以，现在我写的这些。
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q1 The heat input is just q1, Q and we'll define that as capital Q.
输入的热是，我们把它定义为大写的。
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I am going to highlight this by calling this q1, lower case q being the charge here on the nucleus.
我要把它称作q1来进行强调，小写q是指原子核上的电量。
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What is Firm 2's best response as a function of q1?
1的函数即公司2的最佳对策是什么
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So we can just write out what those are, minus w prime over q1 prime q1 is greater than minus w over q1.
因此可以写出，负w一撇除以q1一撇,大于负w除以。
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q1*q2/ That's simply q1, q2 over 4 pi epsilon zero R.
那只是简单的。
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So that says the q1 is the opposite of w1.
这是等温膨胀。
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q1 And it's got q1.
这边的热量是。
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Which means that q1 over T1, that's this delta S thing that we saw before.
也就是说q1除以q2本质上是，我们之前见过的那些含Δ的量。
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Which is to say q1 plus q2 is equal to minus w1 plus w1 prime plus w2 plus w2 prime.
即q1加q2等于2,负的w1加w1一撇1,加w2加w2一撇。
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Q So it's just capital W over capital Q, q1 which is to say it's minus all that stuff over q1.
所以效率就是W除以，等于负的所有这些之和除以。
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Good, so this is our key expression and I'm going to use this expression; I'm going to solve it for q1.
好了,这是个关键表达式,我们一会还要用到它的,我们用它来解出q1
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Q1 I'm going to, just for convenience, call that Q1.
为了方便，称之为。
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So w over q1 prime is greater than minus w over q1.
因此w除以q1一撇大于负w除以q1，等于。
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q1 Or we can write it as one plus q2 over q1.
即1加q2除以。
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The efficiency for the one on the left q1 is minus w over q1, The efficiency prime for the one on the right is minus w prime over q1 prime.
左边的效率是,负w除以，右边的效率是,负w一撇除以q1一撇。
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Let's multiply both sides by 2, I'll get 2q1* is equal to a - c over b - q1*.
等式两边同时乘以2,得到,2q1*=/b-q1
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q2 So I'm going to write this as q2 over q1 over minus one plus q2 over q1.
因此这等于q2除以q1，除以负1加上q2除以,我这么做的目的是。
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Because we've got one q1 and then we're adding q2 over q1 to it, but we've got negative q2 is a positive number.
这是因为这里有个,然后加上了q2除以，但是我们知道负的q2是个正数。
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That says that minus q1 prime plus q1 is greater than zero.
这就是说1,负q1一撇加q1大于零。
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So, in particular, what we're going to do to find out what quantity q1 maximizes this profits for each choice of q2.
所以我们要算出来在不同的q2下,q1取什么值才能最大化利润
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I'll leave the algebra there, transfer it up here, q1* = q2* = a - c over 3b.
先把算式留在那好了,我在这里再抄一遍吧,q1*=q2*=/3b
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I want to do that q2 T1 that q2 over q1 is negative T2 over T1.
因为我知道,除以q1等于负T2除以T1，because，we，know,所以这等于负T2除以。
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Let's just use our expressions q1 that we've found for q2 and q1.
利用q2和q1的表达式,我们有q2除以。
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The way we read this green graph is you give me a choice of Firm 1, q1.
这条绿色的线表示,任意给出公司1的产量q1
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q1 So sufficiency is q1 plus q2 over q1.
所以效率等于q1加q2除以。
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q1 So we saw that efficiency is one plus q2 over q1.
效率等于1加q2除以。
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q2 We can replace this sum by just the some of q1 plus q2.
我门可以用q1加,来替代这些量之和。
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And we wrote something that looks, the energy is equal to minus the Madelung constant times Avogadro's number, 0R0 q1 q2 over 4 pi epsilon zero R zero.
我们写下了，晶格能等于负的马德隆常数,乘以阿伏伽德罗常数，乘以q1q2除以4πε
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e And then q1 is plus qe, q2 is minus e.
也知道q1是+e，q2是。
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