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## Hot answers tagged maxima-minima

7

As others have noted, the minimum and maximum of the $\emptyset$ are undefined, since a minimum or maximum must belong to the set, but the infimum and supremum are well-defined. The infimum of a set is the greatest lower bound. That is, it is the largest number such that no member of the set is smaller than it. Now if we take any real number $x$, it is ...

5

$|x+y| +|x-y|=2$ represents a square of side $2$ centered at the origin and aligned along the axes. $x^2 -6x +y^2 =(x-3)^2 +y^2 -9$ is maximized when $(x-3)^2 +y^2$ is maximized. This is nothing but the square of the distance of $(x,y)$ from $(3,0)$. So the question is, what is the largest possible radius of a circle centered at $(3,0)$ such that it ...

4

$v=x\gamma$ and $x=v\gamma$ give $v=v\gamma^{2}$. If $\gamma \neq \pm 1$ we get $v=0$ and $x=v\gamma =0$. If $\gamma =\pm 1$ the expression becomes $(x+v)^{2}$ or $(x-v)^{2}$. Can you minimize this?

4

We can simplify the problem by treating the function in terms of $x^2$. That is, the maximum value of $\arctan(x^2)/\log(1+x^2)$ is the same as that of $\arctan x/\log(1+x)$ which has derivative $$\frac1{\log^2(1+x)}\left(\frac{\log(1+x)}{1+x^2}-\frac{\arctan x}{1+x}\right).$$ Setting this to zero yields an expression for the stationary point $x_s$ $$\frac{\... 4 My favorite way,$$f(x,y)+20=y^2+y(2x^2-8)+x^4-4x^2-8x+20\begin{align}\Delta_{\text{half}}&=(x^2-4)^2-(x^4-4x^2-8x+20)\\ &=-4(x-1)^2≤0.\end{align}$$This means, f(x,y)+20≥0. Hence, for minimum of f(x,y)+20, we need to take x=1 and y=4-x^2, which gives f(x,y)+20=0. Finally, we deduce that$$\min\left\{f(x,y)+20\right\}=0~ \\ \text {at}~...

3

$\frac {\sin x_1-\sin x_2} {x_1-x_2} \leq 1$ whenveer $x _1 \neq x_2$ by MVT. Also, taking $x_2=0$ and letting $x_1 \to 0$ we see that the expresion takes value as close to $1$ as we want. Hence, the supremum is $1$. To show that this supremum is never attained note that if $x_1>x_2$ and $\frac {\sin x_1-\sin x_2} {x_1-x_2} = 1$ implies that $1=\frac 1{... 3 There are three techniques taught in Calculus for determining whether a critical point is a (local or global) extremum : the second derivative test, the first derivative test, and just evaluation. Here, we find$\frac{\partial f}{\partial x} = 3x^2y^3$, so some critical points are$\{(0,y) \mid y \in \Bbb{R}\}$. Also,$\frac{\partial f}{\partial y} = 3x^3y^...

3

You can find the stationary points: \begin{align} \frac{\partial f}{\partial x}&=4x^3+4xy-8x-8 \\[6px] \frac{\partial f}{\partial y}&=2x^2+2y-8 \end{align} At a critical point $y=4-x^2$ and also $$x^3+x(4-x^2)-2x-2=0$$ that is, $x=1$, that implies $y=3$. Since clearly the function is upper unbounded on the line $y=0$, we just need to show it is ...

2

Your minimization problem is best handled in another set of coordinates. Done right, one can avoid the square roots from the norm term. Firstly, your work appears to tacitly assume that $A^{\mathsf{T}}=A$. I will assume $A$ is symmetric going forward, just for simplicity of notation. If not, then note that since the transpose is the identity on scalars, ...

2

What you've done so far is reasonable. Derivatives won't help, because the variables are required to be integers. Since $\lim_{r\to1}\frac{1+r}{1-r}=\infty$, to get a large value, we want to make $r$ close to $1$, or $a+b+c$ close to $d$. At the same time, we want to make $a+b+c+d$ as large as we can. Let's try $a+b+c-d=1$, which is as small as we can ...

2

$\frac {x+d}{x-d} = \frac {(x-d) + 2d}{x-d} = 1 + \frac {2d}{x-d}$. Now if $x-d = 0$ the whole thing is undefined. Otherwise $x-d \ge 1$ or $x-d < 0$ and $0< d\le 25$. If $x-d < 0$ then as $d > 0$ we have $1 + \frac {2d}{x-d} < 1$. If $x-d \ge 1$ then So $1 +\frac {2d}{x-d} \le 1 + \frac{2d}1 \le 1 + \frac {2\cdot 25}1 = 51$ is the absolute ...

2

Of course. Let $A = \{(x,y): x^2+y^2 \le 1$} $$f: A \to \mathbb{R}$$ $$f(x)=-\sqrt{x^2+y^2}$$ has unique global maximum at $1$ and every point that satisfies $x^2+y^2=1$ is a local minimum. Alternatively, consider the function $|x^2+y^2-1|$. or $(x^2+y^2-1)^2$. Edit after the condition of $\mathbb{R} \to \mathbb{R}$ is imposed. $$f(x)=\begin{cases} 1-x &... 2 For AM-GM, -2xy \le 2|xy| \le x^2+y^2 so 2xy\ge -x^2-y^2.$$5(x^2+y^2)\le 6x^2+2xy+6y^2 =9.$$Then x^2+y^2\le 9/5. The maximum is reached when x=-y=\pm\sqrt{9/10}. You can also get the min value of x^2+y^2. Because 2xy\le 2|xy|\le x^2+y^2$$7(x^2+y^2)\ge 6x^2+2xy+6y^2=9.$$Then x^2+y^2\ge 9/7. The minimum is reached when x=y=\pm\sqrt{9/14}.... 2 Let x= rcostheta , y= rsintheta , put it back into the given relation and get a function of r^2 in terms of theta we need to maximize this r^ 2 wrt to theta ,use calculus or just bounds of trigo function 2 Hint: Find when you can have a stationary point p=(\bar x,\bar y,\bar z,\bar w) \in \mathbb R^4. F:\mathbb R^4\to \mathbb R such that F(x,y,z,w)=h_1(x,y)g_1(z,w)+h_2(x,y)g_2(z,w), where h_i,g_i:\mathbb R^2\to\mathbb R, i=1,2. Supposing h_i,g_i are differentiable, you can impose$$\nabla F=(F'_x,F'_y,F'z,F'_w)=(\underbrace{g_1(z,w)\dfrac{\partial}{...

2

I can get to $4$, by mean value theorem. Proof: Let $x_{\min}$ denote the minimum point. By mean value theorem there is $c_1,c_2$ such that $$f'(c_1)=\frac{f(x_{\min})-f(0)}{x_{\min}},f'(c_2)=\frac{f(1)-f(x_{\min})}{1-x_{\min}}.$$ And then, by mean value theorem there is $c_3$: $$f''(c_3)=\frac{f'(c_2)-f'(c_1)}{c_2-c_1}$$ $$=\frac{\frac{f(1)-f(x_{\min})}{1-... 2 As you've written it, your professor is wrong. However, there is a similar (but definitely different) notion of a greatest lower bound, also called an infimum. This value is defined as the largest value that has no set element larger than it. For an empty set, the greatest lower bound must be \infty. 2 Following Vasya's comment, we observe that the three expressions$$ x_1-x_2 \\ x_1+x_2-x_3 \\ x_1+x_3 are linearly independent, so we can solve (for example) \begin{align} x_1-x_2 \phantom{{}+x_3} & = 0 \\ x_1+x_2-x_3 & = \frac\pi2 \\ x_1\phantom{{}+x_2}+x_3 & = 0 \end{align} to yield x_1 = x_2 = \pi/6, x_3 = -\pi/6 to get the maximum of 5.... 2 As a,c \ge 0 thenP \ge S := ab+bc+ca + \frac{3}{a+b+c}$$We minimize S As a^2+b^2 +c^2 = 3 then$$S =\frac{1}{2} (a+b+c)^2+\frac{3}{a+b+c} -\frac{3}{2}$$Denote x = a+b+c. Because a^2+b^2 +c^2 = 3 then \sqrt{3} \le x \le 3$$S = f(x) := \frac{1}{2}x^2+\frac{3}{x}-\frac{3}{2}$$As f'(x) = x-\frac{3}{x^2}>0 in the interval [\sqrt{3},3] ... 2 As @Aaron Hendrickson already commented, using generalized harmonic numbers$$f(n)=\frac 1 n H_n^{\left(-\frac{1}{n}\right)}$$and the maximum is reached for n=3 for which f(3)=\frac{1}{3} \left(1+\sqrt{2}+\sqrt{3}\right)\sim 1.23406. This is very easily obtained using integer optimization. 2 There is a sign error in your characterization of convex functions via the gradient, it should be$$ \tag{*} f(y)-f(x) \ge \nabla f(x) \cdot (y-x) $$for x, y \in \Bbb R^n. Setting y=0 then gives f(x) \le \nabla f(x) \cdot x, which is the opposite of what we need. But actually equality holds, and that can be seen as follows: If f homogeneous and x \... 2 As mentioned in the comments, we need that f(0) = 0, which follows from the homogeneity. Then$$ 0 = f(0) = f\left( \frac{(-x) + x}{2}\right)\le \frac{f(-x) + f(x)}{2} $$implies f(-x ) \geq -f(x). 1 Generally, you cannot claim this. First, the solution of the differential equations are not necessary a local maximum and a local minimum of the function. They can be either of those or a saddle point. Second you should check the boundary of the function domain: the global maximum and/or global minimum can be there. 1 In general, if f(v,x) achieves local extremum, then \nabla f(x,y) = (\partial_vf,\partial_x f) must be zero vector. So, the partial derivatives that you calculated both need to be set to 0 to find candidates for local extrema. In particular:$$v = \gamma x,\ x = \gamma v \implies (1-\gamma^2) x = 0,\ (1-\gamma^2) v = 0.$$We need to look at different ... 1 Differentiating with respect to either variable loses information. In Leibniz's proofs on the foundations of calculus, he makes one big assumption, and that is that he treats \frac{dy}{dx} as a fraction, and so calculates$$\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}$$at one point. This is impossible for the expression that you have given, because if one ... 1 We have to go over a few cases to analyze your algebraic expression. I have included all possible cases in my answer and I want to point out that I will use only algebra. Using the method of completing the square, we have$$v^2 + x^2 - 2xv\gammax^2-x(2v\gamma)+v^2\begin{align}(x-v\gamma)^2+v^2-v^2\gamma^2≥v^2(1-\gamma^2)\end{align}\begin{...

1

The function does not have a global maximum because $y^4$ grows without bound as $y\to\pm\infty$; the same holds for the $x$ variable. This can be shown by just taking the limits separately, because $f$ splits nicely between $x$ and $y$. However, because of this behavior at infinity and due to its continuity, it must have a global minimum thanks to a variant ...

1

Let set: $\begin{cases}f(x,y)=x^2+y^2\\g(x,y)=6x^2+2xy+6y^2\end{cases}$ The curve $(\mathcal E): g(x,y)=9$ is an ellipse centered at the origin with major axis supported by $y=-x$ and minor axis supported by $y=x$. Indeed if you calculate its equation in the $45^\circ$ rotated basis then you get it in its reduced form: $$g(\tfrac{X+Y}{\sqrt{2}},\tfrac{X-Y}{\... 1 Without AM-GM inequality: Let, x=a+b and y=a-b, then you get$$6(a+b)^2+2(a+b)(a-b)+6(a-b)^2=914a^2+10b^2=9x^2+y^2=(a+b)^2+(a-b)^2=2a^2+2b^2x^2+y^2=2a^2+2 \times \frac{9-14a^2}{10}x^2+y^2=\frac{9-4a^2}{5}≤\frac95\color {gold}{\boxed {\color{black} {\text{max}[x^2+y^2]=\frac 95.}}} If you want to find the minimum value of the ...

1

For $f(x) = x^3 - 3x$ the point $x = 1$ is a local minimizer, but not a global minimizer. The reason for this is that in a small region around $x = 1$, $f(1) =-2$ is the smallest value, meaning it is a local min. The reason it is not a global min is because there are points where the $y$-value is smaller than $-2$. For example, $f(-3) = -18 < -2$, ...

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