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An elementary approach: using $2xy\leq x^2+y^2$, you can write $$xy=\frac{1}{3}(2xy+xy)\le\frac{1}{3}(x^2+y^2+xy)=\frac{1}{3}1=\frac{1}{3}$$ with equality iff $x=y=\frac{1}{\sqrt{3}}$. For minimum, note that $$xy+1=xy+(x^2+y^2+xy)=(x+y)^2\geq 0\implies xy\geq-1$$ where equality realizes when $(x,y)=(1,-1)$ or $(x,y)=(-1,1)$.

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you can do this without calculus. the $x$-coordinates of the points common to both $xy = k$ and $x^2 + y^2 + xy = 1$ satsfies $x^2 + k^2/x^2 + k - 1 = 0$ which can be turned into a quadratic equation for $u = x^2$ as $$u^2 + (k-1)u^2 + k^2 = 0$$ whose discriminant $1 - 2k - 3k^2$ is positive for $-1 \le k \le 1/3.$ so the maximum value of $f$ is $2/3$ ...

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Hint: If you solve the second equation for $y$, you get: $$y = \dfrac{3 x+8}{x+2}$$ Substituting this into the first equation, you get an expression in $x$, as: $$4 x^4+32 x^3+23 x^2-176 x - 256 = 0$$ You can either find the exact roots to this quartic equation or use Netwon's Method or equivalent for numerical results. This gives you two real and two ...

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Let's see. Writing $\nabla f = \lambda \nabla g$, we have $$3 = \lambda . 2x \ \ \text{ and } \ \ 2 = \lambda . 2y$$ The multiplier $\lambda \neq 0$ and hence $x = 3/(2\lambda)$ and $y = 1/\lambda$. Substituting back into the constraint $g$, $\displaystyle \frac{1}{4\lambda^2} ( 9 + 4) = 4$ and we find $\lambda = \pm \sqrt{13}/4$. In the negative case ...

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There is no one Lagrangian that will solve this problem. The minimum could be: inside the cube on one of the faces on one of the edges on one of the corners That's $1 + 6 + 12 + 8 = 27$ different cases. In general, you would have to solve each case, and reject all solutions that are not in the cube or on its surface. But perhaps your $Q$ allows ...

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If you note that $x\neq 0$ and $y\neq 0$ you can divide the second equation by the third and get: $$\frac{2ye^{2xy}}{2xe^{2xy}} = \frac{2\lambda x}{2\lambda y} \ ,$$ which after simplification says that $x^2=y^2$. Taking the first equation, you can then write $$x^2 + y^2 = 2x^2 = 16$$ This gives $x=\pm 2\sqrt{2}$ and therefore four possible solutions ...

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When we use Lagrange method on [a linear combination of concave functions constrained to a linear space] and find a local max, will it in fact then be a global max? Yes: A linear combination of concave functions is also concave. A concave function restricted to a linear space (e.g. the hyperplane $g = c$) is concave. A local maximum of a concave ...

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