Suppose xand y are real numbers and that $x^2 +9y^2 -4x +6y+4=0$ then we have to find the maximum vale of 4x-9y  I found that the equation given is that of ellipse .  I can consider a line $4x-9y=0$ which cut the ellipe .  bt by doing this its getting very comlicated .
122286
Let $x$ and $y$ be real numbers such that $x^2 + 9 y^2-4 x+6 y+4=0$. Find the maximum value of $\displaystyle \frac{4x-9y}{2}$.  My solution: the given function represents an ellipse. Rewriting it, we get $\displaystyle (x-2)^2 + 9(y+\frac{1}{3})^2=1$. To find the maximum of $\displaystyle \frac{4x-9y}{2}$, $x$ should be at its maximum and $y$ at its minimum. Solving the equations, I get that $x = 3$ and $\displaystyle y = -\frac{2}{3}$, but the answer i get is wrong. What am I doing wrong?
43470
I have the folowing exercise (which I've been thinking quite a while and couldn't figure out):  Show that $\forall (x,y)$ in the first quadrant: $$\frac {x^2+y^2}{4}\leq e^{x+y-2}$$  My idea was to work with maxima and minima, but I'm stuck... Any help will be much appreciated!
101787
Let $P(x)=4x^2+6x+4$ and $Q(y)=4y^2-12y+25$.  Find the unique pair of real numbers $(x,y)$ that satisfy $P(x)Q(y)=28$    I can solve this question graphically. $$P(x)Q(y)=28\implies(4x^2+6x+4)(4y^2-12y+25)=28$$ $$4x^2+6x+4=\frac{28}{4y^2-12y+25}$$ I drew the graph of $4x^2+6x+4$ and found it is a upward parabola with minimum value at $(-\frac{3}{4},\frac{7}{4})$ and I drew the graph of $\frac{28}{4x^2-12x+25}$. I found that it is a bell shaped curve whose maximum value occurs at $(\frac{1}{4},\frac{7}{4})$. So I found that the common abscissa is $y=\frac{7}{4}$ which occurs at $x=-\frac{3}{4}$ for the first curve and at $x=\frac{1}{4}$ for the second curve, so solution should be $(-\tfrac{3}{4},\tfrac{1}{4})$  But I do not  know how to solve it algebraically.
26029
Well, I was given a problem, find $x$, if:  $$(x^2+100)^2=(x^3-100)^3$$  I tried everything that I could, I even opened up the brackets which gave an ugly degree 9 equation, I also tried to plot the curves $y=\left(x^2+100\right)^2$ and $y=\left(x^3-100\right)^3$ and locate their point of intersection but it couldn't be done manually.  So, in the end I was forced to use hit and trial after doing which I got the answer, is their any way to solve this algebraically??
47332
Find the global maxima/minima of $f(x,y,z) = x+y+z$ for points inside of $A = \{ (x,y,z) \in \mathbb{R}^3: x^2-y^2 = 1 \wedge 2x+z = 1 \}$   I renamed the conditions of $A$ to a function $g(x,y,z) = x^2-y^2-2x-z = 0$ in order to be able to use Lagrange multipliers.  Deriving $f$: $\nabla(x,y,z) = (1,1,1)$  Deriving $g$:   $g_x = 2x-2$ $g_y = -2y$ $g_z = -1$   By solving the Lagrange system:  \begin{cases} 1 = \lambda(2x-2) \\ 1 = \lambda (-2y) \\ 1 = \lambda (-1) \\ g(x,y) = 0 \end{cases}  I get that $(x,y,z) = (\frac{1}{2}, \frac{1}{2}, -1)$ for $\lambda = -1$. But that point is not inside A.  Is this enough to guarantee that $f|_A$ does not have either maxima nor minima?  (Since the second derivatives are all zero, the second derivative test doesn't give any information.)  Thanks!  EDIT:  By parametrizing $y = \pm \sqrt{x^2-1}$ and $z=1-2x$ and doing the composition with $f$ I get that $f(x, \sqrt{x^2-1}, 1-2x)$ is monotonically increasing and $f(x, -\sqrt{x^2-1}, 1-2x)$ monotonically decreasing. Would that prove that $f$ never reaches either maxima nor minima?
73613
The global maximum and the global minimum of the function $f(x,y,z)=xyz$ with the constraint $x^2+2y^2+3z^2=6$ can be found using Lagrange multipliers.  $\nabla f = \lambda \nabla g$  $g(x,y,z)=x^2+2y^2+3z^2-6=0$  This results into:  $$yz = \lambda 2x$$ $$xz = \lambda 4y$$ $$xy=\lambda6z $$  When looking at the equation $x^2+2y^2+3z^2-6=0$, it can be seen that there are six solutions $(x,y,z)$ with two coordinates equal to zero: $(0,0,\pm \sqrt 2 ),  (0, \pm \sqrt 3, 0 ) ,  (\pm \sqrt 6, 0, 0 ) $  But I got stuck here. Could someone help me with this?
8013
I have the following system of equations:  \begin{eqnarray}\frac{dx}{dt} = x(1 - x^2 - y^2) \\ \frac{dy}{dt} = y(4 - x^2 - y^2) \end{eqnarray}  I want to prove that if a solutions starts (at time $t = 0$) in the region   $$ G = \{(x, y) \in \mathbb{R}^2 : x &gt; 0, y &gt; 0 \text{ and } 1 &lt; x^2 + y^2  &lt; 4\} $$  that it remains there for all $t \geq 0$. I have already proven that, from $G$, a solution cannot cross the $x$-axis where $0 &lt; x &lt; 2$ or the $y$-axis where $0 &lt; y &lt; 2$ (since those line segments are orbits), but the pieces of the circles with radius 1 and 2 (where $x &gt; 0, y &gt; 0$) are the lines I haven't done yet. In this question we can focus on just the circle with radius 1 since the proof for the other circle is completely analogous to it.  In my book, there is a similar problem where they show that if a solution crosses a certain line, that solution attains its maximum or minimum value for $x$ and $y$, and then they derive a contradiction. However, in this case, for $n \geq 1$ the $n$-th derivative of $x$ equals 0 on the circle with radius 1. (If the second derivative were negative, $x$ would attain its maximum on the circle, a contradiction since $\frac{dx}{dt} &lt; 0$ in $G$, so $x$ is decreasing.)  I don't see why a solution can't cross the circle at all. On the circle, $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} &gt; 0$, so after crossing the circle, the solution would go back up to $G$.  Is the circle itself an orbit?  EDIT I think I have found the answer. Using polar coordinates, I found $$ \frac{dr}{dt} = r - r^3 + 3r \sin^2(\theta) $$ Now, $$ \frac{dr}{dt} &gt; 0 \Leftrightarrow 0 &lt; r - r^3 + 3r \sin^2(\theta) \leq r - r^3 + 3r = 4r - r^3 $$ since $0 \leq \sin^2(\theta) \leq 1$. Note that $0 &lt; r \leq 1$ satisfies $4r - r^3 &gt; 0$, and thus $\frac{dr}{dt} &gt; 0$. Therefore, if a solution goes from $G$ to the edge of the unit circle, $r$ must, on the edge, either be decreasing or constant (i.e., its derivative is less than or equal to 0), contradicting the fact that the derivative is positive.  Can someone tell me if this is correct?
115093
I first came across this problem on  and have tried to solve it myself. Here is what I did (may seem laughable to those who know the answer!):  Let $a = x+y$ and $b=1+xy$  I figured that to prove $x+y&lt;1+xy$ to be correct, I need to find an equation which will allow me to plot a graph, $d = b-a$, showing that $b - a$ approaches 0, but never touches it. This is where I am stuck and don't know how to proceed. Presumably, if my method has been correct so far, I need to simplify it in some way so that there is one variable only.   (I haven't really been taught anything like this at school but since I want to study maths later on, I figured it may be helpful to try some challenging questions like this. This seemed fairly easy at first, but can't get my head around it, so any helpful answer which I can learn from is greatly appreciated.)
100892
I have the equation: $$y^2=x^3-2$$ It seems to be deceivingly simple, yet I simply cannot crack it. It is obviously equivalent to finding a perfect cube that is two more than a perfect square, and a brute force check shows no solutions other than $y=5$ and $x=3$ under 10,000.  However, I can't prove it.   Are there other integer solutions to this equation? If so, how many? If not, can you prove that there aren't?   Bonus: What about the more general equation" $$y^2=x^3-c$$ Where $c$ is a positive integer?
72483
The problem is the following:     If $x$ and $y$ are integers such that $\dfrac{4x^2-1}{4x^2-y^2}=k$ is   also an integer, does it implies that $k=1$?   This equation is equivalent to $ky^2+(1-k)4x^2=1$ or to $(k-1)4x^2-ky^2=-1$. The first equation is a pell equation (if $k$ is a perfect square) and the second is a pell type equation (if $k-1$ is a perfect square). I've tried setting several values of $k$ to get some solutions but i got nothing. I'm starting to think that $k$ must be $1$.
118396
How do I find the maximum value for $x\cos {λ}+y\sin {λ}$ for fixed $x,y$ and varying $λ$ ? I would appreciate if only hint is given.
20411
Question:     Given complex numbers $a,b,c$, we have that $|az^2 + bz +c| \leq 1$ holds true for any complex number $z, |z| \leq 1$. Find the maximum value of $|bc|$   It is said this is answer is $$|bc|\le \dfrac{3\sqrt{3}}{16}$$  My idea: let $z=1$,then $$|a+b+c|\le 1$$ let $z=-1$,then $$|a-b+c|\le 1$$ let $z=0$, then $$|c|\le 1$$  let $|\theta|=1$,then we have $$|a(z/\theta)^2+b(z/\theta)+c|\le 1\Longrightarrow |az^2+b\theta z+c\theta^2|\le 1$$ and only this can't solve this problem,Thank you
35665
I have a hyperbola in the first quadrant opening to the right given with the equation $$x=\sqrt{19.8y^2-6.9y+0.7}$$ and I have to find the tangent line from the point $(0, 0.08)$ to the hyperbola (the tangent line with positive slope, so above the hyperbola if it's opening to the right).     The problem is I don't know what the tangent point is on the hyperbola because then I would just be able to find the slope and use $y=mx+b$ to find the equation of the tangent line.  My idea is since there are many lines that intersect the hyperbola from that point I have to maximize the slope between the point and the hyperbola and then I should be able to find the tangent line between $(0, 0.08)$ and the hyperbola. But I don't know how to find the maximum slope between the line that goes through the point and the hyperbola?
52976
The problem I have to solve is:     If tangent lines to ellipse $9x^2+4y^2=36$ intersect the y-axis at point $(0,6)$, find the points of tangency.
69405
If $x&gt;0$, $y&gt;0$, $z&gt;0$ and $x+y+z=6$ then find the minimum value of    $$ f(x,y,z)=\left(x+\frac{1}{y}\right)^2+\left(y+\frac{1}{z}\right)^2+\left(z+\frac{1}{x}\right)^2. $$   Thanks in advance.
122619
e.g. given two numbers $5$ and $6$ the maximum number is $19$, as after $19$ each number can be formed using equation $5n+6m$ by putting different (non-negative) values for $n$ and $m$.     Such number is not possible for $4$ &amp; $6$.     I am not sure how to go-forward. The only thing I have till now is that the number is possible only if the given numbers are pseudo-prime. Any help is appreciated.
86443
We have three points $P=(2,3)$, $Q=(4,-2)$ and $R=(\alpha,0)$. We need to find the value of $\alpha$ such that:    Part 1: $|PR+QR|$ is maximized.  Part 2: $|PR-QR|$ is minimized.  I have  the solution for both. I have understood the solution for case 1 but not for case 2.  Case 1's solution :    Case 2's Solution:   I have understood case 1 completely but in case 2 why was there a need to take the image of point Q? I fail to understand this.