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The solution The answer can be been found on the internet in any number of places. The function $U$ is a Cobb-Douglas utility function. The Cobb-Douglas function is one of the most commonly used utility functions in economics. The demand functions you should get are: $$x(p,I)=\frac{\alpha I}{(\alpha+\beta)p}\qquad y(p,I)=\frac{\beta I}{(\alpha+\beta)q}$$ ...


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The $x$ in the Lagrangian can be any value. However the "subject to" constraints tell us that the domain over which you are to minimize $L$ are only those that satisfy $c^Tx = 0$ and $x^Tx = 1$, i.e. all $x$ in the intersection of the unit $n$-sphere and the plane $c^Tx = 0$. After you have derived the Lagrangian equations $$\frac{\partial L}{\partial \mu} ...


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Imagine that you're aiming to cover as much of the $\sum_i v_i$ square as possible: The bigger the largest inner square, the closer it gets to covering more of the background square. The maximum sum of squares is reached when all but one of the $v_i$ is at the specified minimum - in this case, $1$.


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If there are two numbers with $1\lt v_j\le v_k$, you can decrease $v_j$ and increase $v_k$ to increase the sum of squares. Thus at most one of the $v_i$ is greater than $1$.


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Have you tried a software like Mathematica? There are four solutions to your system: $x=y=Sqrt[5/2]$ and the negative of that root (both with $λ=5/8$), and $x=-y=Sqrt[10]$ and the negative of that root (both with $λ=5/2$).


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You have three equations from the first order conditions: $$2xy^2z^2=2\lambda x \qquad 2x^2yz^2=2\lambda y\qquad 2x^2y^2z=2\lambda z $$ Suppose $x=0$. Then the first equation is satisfied, and the other two equations imply that $\lambda=0$ (since we cannot have $x=y=z=0$ as that does not satisfy the constraint). This gives us infinitely many solutions with ...


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$$\mathrm x^T \mathrm A \mathrm x \geq \lambda_{\min} (\mathrm A) \|\mathrm x\|_2^2 = \lambda_{\min} (\mathrm A) > 0$$ because $\|\mathrm x\|_2 = 1$ and $\mathrm A \succ \mathrm O$. The minimum is attained at the intersection of the eigenspace of $\lambda_{\min} (\mathrm A)$ with the unit Euclidean sphere.


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In the 1D case, the second derivative test is inconclusive when the second derivative vanishes at the critical point. You can see this by looking at $f(x)=x^3,g(x)=x^4,x_0=0$; the second derivative test gives the same information but the behavior is different. The analogous thing happens in higher dimensions, including in constrained problems. tl;dr: You ...


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Say you have: $$\left< 3z, 6, 3x \right> = \lambda \left< 2x, 4y, 2z\right>$$ Clearly this is equal to: $$\left< z, 2, x \right> = \frac{2\lambda}{3} \left< x, 2y, z\right>$$ You can let $\lambda^* = 2\lambda/3$ and proceed as usual. As you said, the direction is the most important part of Lagrange Multipliers, for the most part $\...


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If $I=px+qy$, then $y = (I-px)/q$, so $x^ay^b =x^a((I-px)/q)^b =x^a(I-px)^b/q^b $. Differentiating, we want $\begin{array}\\ 0 &=(x^a(I-px)^b)'\\ &=ax^{a-1}(I-px)^b-x^apb(I-px)^{b-1}\\ &=x^{a-1}(I-px)^{b-1}(a(I-px)-xpb)\\ &=x^{a-1}(I-px)^{b-1}(aI-apx-xpb)\\ &=x^{a-1}(I-px)^{b-1}(aI-xp(a+b))\\ \text{so}\\ x &=\dfrac{aI}{p(a+b)}\\ \...


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You can certainly simplify that system! $$\lambda = \frac{\alpha x^{\alpha - 1}y^\beta} p$$ $$\lambda = \frac{\beta y^{\beta - 1}x^\alpha} q$$ Thus $$\frac{\alpha x^{\alpha - 1}y^\beta} p = \frac{\beta y^{\beta - 1}x^\alpha} q$$ and $$q\left(\alpha x^{\alpha - 1}y^\beta\right) = p\left(\beta y^{\beta - 1}x^\alpha\right)$$ You can reduce powers: $$q\...


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The Hessian of a function is a generalization of the 2nd derivative of a single-variable function $f(x)$. If $x=a$ is a critical point of $f(x)$, a sufficient criterion for $a$ be a maximum is that the second derivative of $f$ at $a$ be negative: $f''(a) < 0$ (https://en.wikipedia.org/wiki/Derivative_test#Second_derivative_test). For multivariable ...


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Unless you restrict the problem to $n \geq 1$ the statement is false. Consider $x=1, y=2, n=\frac12$: $$ \sqrt{2} < \frac32 \implies 1+2+2\sqrt{2} < 6 \implies (1+\sqrt{2})^2 < \sqrt{6}^2\\ \implies 1+\sqrt{2}<\sqrt{6} \implies \frac{1+\sqrt{2}}{2} < \sqrt{\frac{3}{2}} \implies \frac{1^n+2^n}{2} < \left( \frac{1+2}{2}\right)^n $$



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