Find the point on the curve farthest from the line $x-y=0$. the curve $x^3-y^3=1$ is asymptote to the line $x-y=0$. Find the point on the curve farthest from the line $x-y=0$.can someone please explain it to me what the question is demanding?
I cant think it geometrically as I am not able to plot it
Also is there any software to plot such graphs.?
 A: NOTE: you do need to know that the distance of a point $(x,y)$ from the line $y=x$ is  $|x-y|$ (multiplied by the constant $1/\sqrt 2$). There are many ways to figure that out, from many viewpoints.
ORIGINAL: If we write
$$  u = x-y, $$
$$ v = x+y,  $$
we can use the identity
$$ x^2 + xy + y^2 = \frac{1}{4} \left(  (x-y)^2 + 3 (x+y)^2  \right)  $$
to find
$$ x^3 - y^3 = \frac{1}{4} \; u \; \left(  u^2 + 3 v^2  \right),  $$
so the curve becomes
$$  u \; \left(  u^2 + 3 v^2  \right) = 4 $$
May be worth emphasizing that, since we must have $u > 0,$ we always have
$$ u^3 = 4 - 3 u v^2 \leq 4.  $$
Therefore $u$  achieves its maximum when $v=0,$ at which point
$$  u^3 = 4. $$
This relates to a point where $y=-x,$ so $u=2x$ there and the value of $x$ satisfies
$$  8 x^3 = 4.  $$
PLOT:

A: Rotate by $-\pi/4$. This corresponds to $x\mapsto x-y$ and $y\mapsto x+y$ (up to a scalar factor). Then your line becomes $x-y-(x+y)=0\iff y=0$ and the equation of the curve becomes
$$(x-y)^3-(x+y)^3=1\iff -2y^3-6yx^2=1\iff x^2=-\frac{1+2y^3}{6y}$$
as $y$ can't be $0$. You want to find the farthest point from the line $y=0$, so you need to minimize $y$ so that the fraction on the right is positive. So the solution is $(0,-1/\sqrt[3]2)$, which is
$$(1/\sqrt[3]2,-1/\sqrt[3]2)$$ in the original coordinates.
A: Since I don't have a good-enough ( or high -enough ) reputation, I will write my thoughts here:
You have two sets of points in the plane, $S_1=(x,x)$ and EDIT $S_2=(x,(x^3-1)^{1/3})  $
Now use the Euclidean distance and minimize the distance.
A: What the question is demanding is this: the curve gets closer and closer to the line; so it makes sense that there’s a point on the curve that’s farthest from the line.
I’ll show you two ways of doing this: the first is special, the second is general.
The first is to notice, once you’ve drawn the picture, that both $x^3-y^3=1$ and $x-y=0$ are symmetric about the line $y=-x$. the explicit description of the reflection is $(x,y)\mapsto(-y,-x)$. It makes sense that the one point on the curve that’s fixed with respect to this symmetry will have all sorts of special properties. Maybe a local minimum of distance to the line, but much more likely a maximum of this distance. With your picture, you see that it must be a maximum. What is this point? It’s the point on the curve that’s on the line of symmetry $y=-x$. So you make that substitution, get $2x^3=1$, $x=1/\root3\of2$, so the point is $(1/\root3\of2,-1/\root3\of2)$.
The other, general, method, is to use the fact that the closest and farthest points to a straight line have the property that their tangent is parallel to the line (assuming your curve is differentiable, of course). This is a generalization of the fact that the points on a graph $y=f(x)$ that are closest and farthest from the $x$-axis have tangents that are parallel to the $x$-axis (i.e. $f'(x)=0$ there). So you just want a point on your curve where the tangent has slope $1$, same as your line.
That’s easy: differentiating implicitly, you get $3x^2dx-3y^2dy=0$, $dy/dx=x^2/y^2$, and for this quantity to be equal to $1$, you need either $x=y$ or $x=-y$, The former is impossible on your curve, so $x=-y$, and you get the same values for the coordinates as above.
You can do it with Lagrange Multipliers, too, but it boils down to the “general” method I just showed you.
A: An examination of plot in this case shows that you need only to find intersection point $ (x_1,y_1) $ for  the curve and its orthogonal curve through origin:
$$ x^3 - y^3 =1,\, x + y = 0 $$
The maximum distance is then $\sqrt{x_1^2 + y_1^2}.   $
