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3

The pre-order being complete has no topological meaning, but purely set-theoretic. It means that for any two points $x,y$ in the domain of the pre-order, we must have $x \succsim y$ or $y \succsim x$ (or possibly both here, because we have a pre-order, so we can have both at the same time (invariant goods (?), or some such thing, economics is not my field, ...


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Your textbook isn't wrong. $N$ is a fixed number (6 in the example), so the sum is over a finite number of terms, which is perfectly ok. Besides, the sum isn't the harmonic series... It is $$\frac{1}{N}+ \frac{1}{N}+ ... + \frac{1}{N}$$ where there are $N$ summands. So in the example $$ \frac{1}{6}+ \frac{1}{6}+ \frac{1}{6}+ \frac{1}{6}+ \frac{1}{6}+ \...


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Use Lagrange Multipliers. $\mathcal{L}=50L^{0.2}K^{0.8}-\lambda (1000-L-K)$ Take first order condition with respect to K and set it to zero: $50\times0.8\times L^{0.2}K^{-0.2}=-\lambda$ Take first order condition with respect to L and set it to zero: $50\times0.2\times L^{-0.8}K^{0.8}=-\lambda$ Take first order condition with respect to $\lambda$ ...


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Instead of payout, think in terms of the slope of the payout (i.e the delta). If you long a call at $A$ and cover it by shorting a call at $B > A$, the slope will be $1$ between $A$ and $B$ and $0$ otherwise. If you long a put at $D$ and cover it by shorting a put at $C < D$, the slope will be $-1$ between $C$ and $D$ and $0$ otherwise. In the ...


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The inequality is false. Take $K=2$, $\gamma=1/2$ and $$ C_1=100,\quad C_2=1,\quad N_1=1,\quad N_2=100. $$ Then $$ \frac{\sum_{i=1}C_{i}}{\sum_{i=1} {N_i}^\gamma {C_{i}}^{1-\gamma}}=\frac{101}{20}>5 $$ and $$ \frac{K\max_iC_{i}}{\sum_{i=1} {N_i}^\gamma \max_k\{{C_{k}}\}^{1-\gamma}}=\frac{200}{10\cdot(10+1)}<2 $$ PS. A commentator correctly remarked ...


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You have $p+7=\sqrt{\frac{4000}{Q}}$. Increasing $Q$ by 1 decreases the price that can be charged (on all sales) by $\Delta p$, where $p-\Delta p+7=\sqrt{\frac{4000}{Q+1}}$. So we have $\Delta p=\sqrt\frac{4000}{Q}-\sqrt{\frac{4000}{Q+1}}=\sqrt{\frac{4000}{Q}}\left(1-\sqrt{\frac{Q}{Q+1}}\right)$ $=(p+7)\left(1-\sqrt{\frac{1}{1+\frac{1}{Q}}}\right)$. We now ...


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The production function states the quantity that a firm can produce. So if it produces $10$ units: $$10=f(L,M)=L^{1/2}M^{1/2}$$ Hence: $$100=LM$$ $$\frac{100}{L}=M$$ We know the cost will be: $$C=9L+81M$$ $$=9L+\frac{8100}{L}$$ Minimizing this using standard procedures $C'(L)=0$.. gives $L=30$ $C=540$ so I think you typed something wrong. I have ...


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(1) does not imply (2), and (1) and (2) together do not imply (3). There are many non-dictatorial methods that fail to satisfy (2); see this Wikipedia article. For a simple method that satisfies (1) and (2) but not three, let $u$ and $v$ be two designated voters. If $u$ and $v$ have the same preferences, their preferences are adopted by society; if not, ...


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First we translate: $$\frac{dP}{dt}=-\frac{1}{2}(S(P)-D(p))$$ Then we substitute: $$\frac{dP}{dt}=-\frac{1}{2}(5P-60) $$ But now you have a first order differential equation. hint: You can seperate them. you're going to have something involving $e$, and something constant, involving your initial condition. Here are some notes


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First step is to solve for the rate of $p$, which is $p'$. $$S-D=80+3p-(140-2p)=5p-60$$$$\frac{S-D}{2}=\frac{5p-60}{2}$$ Since $p$ is decreasing at the rate, $p'$ needs to be negative.So$$p'=\frac{60-5p}{2}$$ Now it's easy to get $$2p'+5p=60$$It's a non-homogeneous first-order differential equation. $$2D+5=0,D=-\frac52$$Therefore you get the general solution$...


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Let's phrase the problem mathematically. You are being asked to solve:: $\frac{dp}{dt} = -\frac{1}{2} (S(p)-D(t))$ So substituting in the given supply and demand equations: $\frac{dp}{dt} = -\frac{1}{2} (5p-60)$ Separating the variables: $\frac{dp}{p-12} = -\frac{5}{2} dt$ Integrating both sides, we get the general result: $\ln|p-12| = -\frac{5}{2}t+C$...


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You appear in effect to be assuming that $\succsim$ is the same as ordinary $\ge$. This need not be the case. In fact, the whole point of the argument is that the lower contour sets for any continuous preorder on $X$ are closed, so that in this sense all continuous preorders on $X$ behave like the familiar natural order $\ge$. Of course the steps of the ...


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It seems that posers of these problems sometimes delight in expressing the marginal functions in terms of $ \ p \ $ , rather than in terms of $ \ q \ $ which is used in the definitions. On the other hand, it is a reasonable way to do things frequently, since the business has direct control (generally) over the price set, but not over the demand. One way to ...


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You can look up the mathematical definitions in any game theory text book. The interpretation is basically that the players' actions reinforce/offset (complement/substitute) each other. For example, if a buyer is more likely to buy a good at some price, when it is more likely that all other buyers are taking that price, then buyers are strategic complements....


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As $x,y$ goes to infinity, profit goes to negative infinity. So, the maximum exists. We need to check the critical values. As economics only cares about positive numbers, critical values are $x=0,y=0$ and the values that make the partial derivatives $0$. For $x=0$, the profit is $$p(y)=1200-10y^2+96y$$ $p'(y)=96-20y$, thus, $p$ will be maximum at $y_0=24/5$....



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