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Hint You are asked to maximize $F=x^2 y$ subject to the constraint $G=x^2+y^2-3=0$. So, using Lagrange multipliers, let us minimize $$\Phi=x^2 y+\lambda (x^2+y^2-3)$$ The derivatives are $$\Phi'_x=2 \lambda x+2 x y=0$$ $$\Phi'_y=x^2+2 \lambda y=0$$ $$\Phi'_{\lambda}=x^2+y^2-3=0$$ The solutions of this system are $$x=0,y=-\sqrt 3,\lambda=0$$ $$x=0,y=\sqrt ... 1 Here is a plot that shows what's going on geometrically with Lagrangian multipliers in this case. The red line is the x^2+y^2=3 constraint, whereas the underlying contour plot is f(x,y)=x^2 y (lighter regions are at higher values: \hspace{3 cm} The key observation is that the contour which just 'kisses' the constraint line represents larger values ... 0 Alternative method:$$v=\sin(\theta)f = 3 \cos(\theta)^2 \sqrt{3} \sin(\theta)=3\sqrt{3} v(1-v^2)f' = 3\sqrt{3}(1-3v^2)$$Hence when f is maximised v= \pm 1/\sqrt{3}, and we can see from the expression of f this occurs at the positive value of v, leading to a maximum for f of 2. Some trigonometry remains to get the full details. 1 I had made errors in my partial derivatives of g: They should be ∂g/∂x = 2xy ∂g/∂y = 2y Making the solution alot easier Thankyou to @semiclassical for the help 3 Your description of Lagrange multipliers, and of the alternative quadratic penalty method, is a bit off. Rather than jointly minimizing over (x,y,\lambda) \in \mathbb{R}^3, we typically want to fix \lambda\in\mathbb{R} and then minimize over (x,y) \in \mathbb{R}^2. The solution will be parameterized by \lambda, so we then appropriately choose ... 0 The problem may be simplified by working with Y = XX^{\top}, since$$ \text{trace}(X^{\top}UU^{\top}XX^{\top}UU^{\top}X) = \text{trace}( U^{\top} YU U^{\top} YU). $$Then write Y = VSV^{\top}, where S is D \times D diagonal with non-negative non-increasing diagonal entries s_j and V is D \times D orthogonal. Setting \tilde U = V^{\top} U, ... 1 For sure, there is another way if you look for a solution which does not use Lagrange multipliers (I should consider it simpler). You want to minimize$$\Phi= \sum_i (f(x_i)-y_i)^2$$using$$f(x_i)=\beta_0+\beta_1 x_i +\beta_2 x_i^2$$and you want to constraint the model such that f(x_1)=y_1. The constraint allows you to eliminate \beta_0 and let you ... 0 You have an inequality as a constraint. So I would use the Kuhn-Tucker conditions (KKT). But you have an equalility, too. It can be split as follows: g_i(x)=0 \Leftrightarrow \begin{cases} g_i(x) \leq 0 \\ \\ g_i(x) \geq 0 \end{cases} \Leftrightarrow \begin{cases} g_i(x) \leq 0 \\ \\- g_i(x) \leq 0 \end{cases}  max \ \mathcal L=-\sum_{i=1}^n ... 0 The problem of least squares is:$$\min_\beta \sum_i (f(x_i)-y_i)^2$$Where f(x_i) is your favorite form, in your case f(x)=\beta_0+\beta_1 x +\beta_2 x^2 Then what do you want is the next:$$\min_\beta \sum_i (f(x_i)-y_i)^2\text{s.t.} f(x_1)=y_1$$What is equivalent to minimize the lagrangian:$$L(\beta,\lambda)= \sum_i ...

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You found the minimum area (if you accept negative area), which is one of the extreme solutions. What about $\lambda = -\frac{1}{4}$, though? You concluded from $\frac{1}{4\lambda^2}=4$ that $\lambda = \frac{1}{2}$ (I think you meant $\lambda = \frac{1}{4}$), but there are two roots. Also, are you sure that you are solving the problem you want to be ...

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The area is wrong. For a triangle inscribed in a circle (WLOG), you can assume one axis is parallel to the $x$ axis, say at height $y$. Then, the three points are: $$\left(\sqrt{4-y^2},y\right),\ \left(-\sqrt{4-y^2},y\right), (x_3, y_3)$$ The area is therefore: $$A=(y_3-y)\sqrt{4-y^2}$$

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The function we want to maximize $f(x,y)=2xy$, the constraint function is $g(x,y)=y+x^2-4=0$. The condition that must be satisfied is $\nabla f=-\lambda\nabla g$. This gives us two equations: $$\left(\begin{array}\\2y\\2x \end{array}\right)=-\lambda\left(\begin{array}\\2x\\1 \end{array}\right)$$ $$y=-\lambda x ; 2x=-\lambda$$ $$\lambda=-\frac{y}{x}$$ ...

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If you want to use Lagrange multipliers for this problem, your function is $$F=x y+a(y-4+x^2)$$ Now compute the partial derivatives and impose them to be equal to zero $$F'_x=2ax+y=0$$ $$F'_y=a+x=0$$ $$F'_a=y-4+x^2=0$$ Using successive eliminations leads to two possible solutions $$\left\{\left\{x\to -\frac{2}{\sqrt{3}},y\to \frac{8}{3},a\to ... 1 From$$(y,x)=\lambda(2x,1)$$you get (x,y)=(\lambda,2\lambda^2). Since this point must lie on the parabola, y=4-x^2 must hold, so:$$ 2\lambda^2 = 4-\lambda^2,\quad \lambda^2=\frac{4}{3}$$and the maximum value for the area is 2xy=4\lambda^3=\frac{32}{9}\sqrt{3}. We double-check this. The maximum for$$ g(x) = x(4-x^2) $$is attained when ... 1 My suggestion is a bit different: Show that the maximum is not going to change if you expand the set to be the intersection of the positive orthant and the unit ball ( a convex set ) Show that if a solution (a,b,c,d) is such that a\neq b the we have other solution(s) that can be obtained by permutation of the entries, for example, (b,a,c,d). Show ... 1 Your sum is a sum of geometric means of x_1^3, x_2^3, x_3^3,x_4^3 you know each is less than the arithmetic means so the max, M, satisfies$$M\le x_1^3+x_2^3+x_3^3+x_4^3=f(x_1,x_2,x_3,x_4) On the sphere, $f$ has a maximum at $(x_1^2,x_2^2,x_3^2,x_4^2)=\lambda(x_1,x_2,x_3,x_4)$ implying $x_i=x_j$ for all $i,j$. Since the AM is maximized there and ...

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