So say I have an ellipse defined like this:  $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$  I have to find the largest possible area of an inscribed rectangle.  So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefine $y$ in terms of $x$:  $$\frac{y^2}{4} = 1 - \frac{x^2}{9}$$ $$y^2 = 4 - \frac{4x^2}{9}$$ $$y = \sqrt{4 - \frac{4x^2}{9}}$$  So the area function is now:  $$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}$$ $$A' = \frac{4x}{2 \cdot \frac{-8x}{9}} + \sqrt{4 - \frac{4x^2}{9}} \cdot 4$$  Is this the right track? Was there something simpler I could have done? this looks gnarly? Can someone help me finish this up?  So this track seems to difficult, another approach. Can I square the area first, find the derivative of that to solve for x?  So the Area = $4x \cdot \sqrt{4 - \frac{4x^2}{9}}$  Is this valid?  $$Area^2 = 16x^2 \cdot (4 - \frac{4x^2}{9}$$  $$= 64x^2 - \frac{64x^4}{9}$$  Derivative:   $$ \frac{d}{dx} Area^2 = 128x - \frac{256x^3}{9}$$ $$128x(1-\frac{2x^2}{9}$$  So critical values: $x = 0, \frac{3}{\sqrt{2}}$  because the derivative equals 0 when:  $$2x^2 = 9$$ $$x = \frac{3}{\sqrt{2}}$$  Plugging this value of x into y we get that $y = \sqrt{2}$ so the Area is 3.  Is this valid? If so why? Does squaring not cause any problems?
11864
The actual problem reads:      Find the area of the largest rectangle that can be inscribed in the ellipse    $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$   I got as far as coming up with the equation for the area to be $A=4xy$ but then when trying to find the derivative I don't think I'm doing it right.
116593
A circle has the same center as an ellipse and passes through the foci $F_1$ and $F_2$ of the ellipse,such that the two curves intersect in $4$ points.Let $P$ be any one of their point of intersection.If the major axis of the ellipse is $17$ and the area of the triangle $PF_1F_2$ is $30$,then find the distance between the foci.    Let the center of the ellipse and the circle be $(0,0)$ We are given $2a=$length of major axis$=17$. Let the coordinates of foci be $F_1(c,0)$ and $F_2(-c,0)$ We need to find $2c$. Area of $PF_1F_2=\frac{1}{2}\times 2c\times$perpendicular distance between $P$ and the axis of the major axis of the ellipse. I do not know how to solve it further.
117638
In the following diagram, let the side length of the large square be $s$.    How do I specifically determine the numerical value of the area of the blue region?  I can easily show the equivalence between the blue and yellow area, but I don't think this helps determine the area. If possible, I would like to have a method without integration.
29150
In this wikipedia, article  its stated that the circle is the closed curve which has the maximum area for a given arc length. First, of all, I would like to see different proofs, for this result. (If there are any elementary ones!)  One, interesting observation, which one can think while seeing this problem, is: How does one propose such type of problem? Does, anyone take all closed curves, and calculate their area to come this conclusion? I don't think that's the right intuition.
98478
I have a programming problem and I've forgotten some of my math rectangle formulas. I have to make a program that creates a rectangle that isn't parallel to the X and Y axis, I assume this means a diagonal rectangle in the first quadrant of a Cartesian graph.  My inputs are the following:    coordinates of the vertex closest to the X axis and the origin angle the long side makes with relation to the X axis. area of the rectangle length of the long side   The outputs expected are the following:   length of the short side,  coordinates of all the rest of the rectangle's vertices.   Given these input how do I calculate the output mentioned?
24707
A square with edge length 2 cm has semicircles drawn on each side. Find the total area of the shaded region.   Here is an image of the diagram shown :       Please show your work in pictures, numbers, words, anything. (Try to keep it to a Grade 8 level too)
94770
For an $n$-dimensional ellipsoid in $\mathbb{R}^n$ centered at $\mathbf{0}_n$, defined by a set of vectors $\mathbf{v}_i$ giving the directions of the principle axes and $\lambda_i$ representing the corresponding magnitudes of each axis (for $i=1,\ldots, n$), is there a straightforward way to work out the maximal $x$-coordinate, for example, on the ellipsoid?  $\textbf{2D case:}$    For the 2D case let $\mathbf{v}_1=\left(x_1,\;y_1\right)^T$ with $\lambda_1=a$ represent the major axis and $\mathbf{v}_2=\left(x_2,\;y_2\right)^T$ with $\lambda_2=b$ represent the minor axis.  Here the geometry is not too complicated.  We can define a parametric equation for the x-coordinate of the curve in terms of an angle $t$ as $$x=a\cos(t)\cos(c)-b\sin(t)\sin(c)$$ which with some basic calc and trig can be shown to be maximized at $$ x_\text{max} = \frac{a^2\cos(c)^2+b^2-b^2\cos(c)^2}{a\cos(c)\sqrt{\frac{a^2\cos(c)^2+b^2-b^2\cos(c)^2}{a^2\cos(c)^2}}} $$  Similarly a maximal point for $y$ can also be obtained.  $\textbf{General case:}$  For the general case, however, I am stumped. I imagine an analytic result is pretty difficult, so any advice on a numerical approach for finding such a solution in general would also be appreciated.
74792
Using Lagrange multipliers, it can be shown that a triangle with given perimeter has the maximum possible area, if it is equilateral. Is there a simple geometric proof of that fact ?
106132
For a hobby project gone too far, I'm trying to find the maximum separation between two ellipses. I have only a high-school-graduate-level grasp of algebra and calculus, and am happy to dig deeper than that... but I need some guidance.  I have 3 functions of the same form, $X(p, q)$, $Y(p, q)$, and $Z(p, q)$ for which I am trying to evaluate $p$ and $q$. I know that $X+Y+Z=0$. I know I need to express p and q in terms of each other, but as you can see below, the three functions together are quite complex. I've been futzing around with trig identities to try and isolate either variable, but am stuck. Perhaps fresh eyes will help; can anyone here see a way forward?  All values are known, except $p$ and $q$.  $X(p,q) = [b_1V_{x1}\cos(p)-a_1U_{x1}\sin(p)][(C_{x2}+a_2U_{x2}\cos(q)+b_2V_{x2}\sin(q))-(C_{x1}+a_1U_{x1}\cos(p)+b_1V_{x1}\sin(p))]$  $Y(p,q) = [b_1V_{y1}\cos(p)-a_1U_{y1}\sin(p)][(C_{y2}+a_2U_{y2}\cos(q)+b_2V_{y2}\sin(q))-(C_{y1}+a_1U_{y1}\cos(p)+b_1V_{y1}\sin(p))]$  $Z(p,q) = [b_1V_{z1}\cos(p)-a_1U_{z1}\sin(p)][(C_{z2}+a_2U_{z2}\cos(q)+b_2V_{z2}\sin(q))-(C_{z1}+a_1U_{z1}\cos(p)+b_1V_{z1}\sin(p))]$  Edit - A couple threads I've been trying to springboard off of:        I arrived here by taking the partial derivatives of the distance function for two points, one on each ellipse. Once I can find the critical points from that, I should have my solution(s).   The distance function is from the ellipses' parametric equations:  $x(θ)=C_x+aU_x\cos(θ)+bV_x\sin(θ)$  $y(θ)=C_y+aU_y\cos(θ)+bV_y\sin(θ)$  $z(θ)=C_z+aU_z\cos(θ)+bV_z\sin(θ)$  (Where $C$ is the centrepoint of the ellipse, $a$ and $b$ are its semimajor and semiminor axes, respectively, and $U$ and $V$ are unit vectors defining the plane of the ellipse. Angle $θ$ describes position on the ellipse)  So the distance function is:  $$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$$  As the derivative of $\sqrt{n}$ is ${n'\over2\sqrt{n}}$, I only need concern myself with where the numerator ($n'$) is equal to $0$:  $$n'=-2f'(p)[-f(p)+j(q)]-2g'(p)[-g(p)+k(q)]-2h'(p)[-h(p)+l(q)]$$  (Where functions $f$, $g$, and $h$ refer to the parametrics for the 1st ellipse, and functions $j$, $k$, and  $l$ to those for the 2nd. $p$ and $q$ are used instead of $θ$)  $n'=0$ is essentially of the form $0=X+Y+Z$ as outlined above.
103591
I've seen the geometric argument for why any differentiable function $f(x)$ gives the rate of change of the area under its own curve to $x$ for a specific input $x$, and it makes sense to me.  It also makes sense that if $f(x)$ gives you $\frac{dA}{dx}$, then $F(x)$, its antiderivative, should give you the actual area under the curve of $f(x)$.  What I don't understand however, is why does $F(x)$ give you the area from $\textbf{0}$ specifically to $x$ given a single input $x$? In other words, where does that $0$ come from?  Because $f(x)$ gives you $\frac{dA}{dx}$ at $x$ regardless of how much area comes "before" (for smaller values of $x$).   How do we decide that $F(x)$ starts computing the area under the curve of $f(x)$ from $0$ and not some other starting point? Obviously it's very convenient but it seems pretty arbitrary to me.
20504
Suppose, for the sake of illustration, that $AF=2$, where $F\mathbb{'s}$ coordinates are $(2,0)$ and $A$ is the point of origin; $E=(0,1)$ and circle $A$ has $R=1.5$  How do I solve the area of the intersection of the circle and rectangle by integration?  If we let $f(x)=1$, then the area of the rectangle is defined by: $$\int_0^Ff(x)dx$$   If we let $g(x)=\sqrt{R^2-x^2}$, then the area of the circle is defined by: $$4\int_0^Rg(x)dx$$ Now my best attempt to find the area is through: $$\int_0^Rg(x)dx-\int_0^{\sqrt{R^2-1}}(g(x)-1)dx$$  The first integral gives me a quarter of the area circle $R$, the second gives me the area of the circle that is outside the rectangle, because the circle and the top line intersects at $x=\sqrt{R^2-1}$  Is my solution correct?
109848
If $ \frac {x^2}{a^2} + \frac {y^2}{b^2} = r^2$ is an ellipse, with the parameterization $x(θ)≔r(a \cos ⁡θ,b \sin⁡ θ ),$ I have to find the value of $θ$ giving the minimum distance from $P(p,q)$ (not on the ellipse) to the ellipse is given by a quartic in $t= \tan⁡( \frac {θ}{2}).$ A necessary condition for $x$ to be the closest point to $P$ is that $P-x$ is perpendicular to the tangent vector in $x ,$ i.e. $(P-x(θ) ). x' (θ)=0$  I can't handle the above condition to make an fuction (e.g. $f(θ)$) to find the minimum value by calculating the rerivative $f'(θ)=0$ for example. Then I have to prove that rational, non-zero values of $a, b, p, q$ can be found such that the quartic factorises as the product of two quadratics with rational coefficients. Any help? Thank you
69336
Can someone check this for me, please? The exercise is just to find a expression to the nth derivative of $f(x) = e^x \cdot \sin x$. I have done the following:  Write $\sin x = \dfrac{e^{ix} - e^{-ix}}{2i}$, then we have $f(x) = \dfrac{1}{2i} \cdot (e^{(1+i)x} - e^{(1-i)x})$.   Taking the derivatives: $f^{(n)}(x) = \dfrac{1}{2i} \cdot ((1+i)^n e^{(1+i)x} - (1-i)^n e^{(1-i)x})$  Now, I use that: $$ (1+i)^n = {\sqrt{2}}^n \cdot \left(\cos\dfrac{n \pi}{4} + i \sin\dfrac{n \pi}{4}\right) \\(1 - i)^n = \sqrt{2}^n \cdot \left( \cos \dfrac{-n \pi}{4} + i \sin \dfrac{-n \pi}{4} \right)$$  Plugging that mess, I get: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot \left(\left(\cos \dfrac{n \pi}{4} + i \sin \dfrac{n \pi}{4}\right) e^{ix} - \left(\cos \dfrac{-n \pi}{4} + i \sin \dfrac{- n \pi}{4}\right) e^{-ix} \right)$$  But, $e^{ix} = \cos x + i \sin x$, and using Moivre's theorem, that makes: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot \left(\cos \left( x + \frac{n \pi}{4}\right) + i \sin \left( x + \frac{n \pi}{4}\right) - \left(\cos \left( - x  - \frac{n \pi}{4}\right) + i \sin \left( -x -\frac{ n \pi}{4}\right)\right)\right)$$  and since $\cos$ is an even function, and $\sin$ is odd, we get: $$f^{(n)}(x) = \dfrac{e^x}{2i} \sqrt{2}^n \cdot 2i \sin \left(x + \dfrac{n \pi}{4}\right)$$  Simplifying, the answer would be $f^{(n)}(x) = e^x \cdot \sqrt{2}^n \cdot \sin\left(x + \dfrac{n \pi}{4}\right)$. I'm almost positive that this is it, but I just want to be sure. Thank you in advance!
49654
How can I find the maximum area possible of equilateral triangle that inside a square whose sides have length a. And how does that triangle look like? Can we construct it (with compass and straightedge)?
44900
I wish to calculate the volumes of a truncated cone whith asymetry over all axes and ellipses as base and top - how do I do that?  I have height and radii of the corresponding ellipses. (I assume correctly I need the geometric means of the 2 radii describing the ellipsis.)  Kind Regards
56956
I wondered if the centroid of an area could be derived as a limit from the centroid of the solid of revolution built by the same area revolving around an axis of revolution.   e.g.: the  is $C_{V} = \frac{3\pi}{16}r\frac{\sin[\theta]}{\theta}$, where $r$ is the radius of the sphere and $\theta$ is the dihedral angle. The  is $C_{A} = \frac{4r}{3\pi}$  Now I tought it was possible a limit procedure like $\lim\limits_{\theta\to 0} C_{V}$=$C_{A}$, but that's false and I would like to know why (I guess that has something to do with the passage from three-dimension to two-dimension)  Thanks
102651
There are 2 questions(part of same question but I divided it into two):    Q1. Prove that the length of the focal chord of the ellipse $\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$ which is inclined to the major axis at an angle $\theta$ is $$\frac {2ab^2}{a^2\sin^2\theta+b^2\cos^2\theta}$$ I tried to solve this using the parametric form of a line, i.e., $(x,y)=(ae+r\cos\theta,r\sin\theta)$, plugging this into the given equation to find $r_1-r_2$ which is giving a different solution.    Q2. If $PSQ$ and $PS'R$ are two chords of an ellipse through its two focii S and S', then prove that $$\frac {PS}{SQ}+\frac {PS'}{S'R}=2\left(\frac {1+e^2}{1-e^2}\right)$$ Here I fixed $P \equiv (a\cos\theta,b\sin\theta)$ and found out the parametric angles of $Q$ and $R$ using the result $$\tan\theta_1/2*\tan\theta_2/2=\left(\frac {e-1}{e+1}\right)^{\pm1}$$, $+$ when chord passes through $S$ and $-$ when chord passes through $S'$. But it gives a very ugly expression. So how do I do it  in an easy way? Hints are welcome.