Mixing
How do I find the maximum perimeter of a rectangle inscribed in an ellipse?
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
The problem I've been stuck on is this:
A rectangle is inscribed in the ellipse $$fracx^220 + fracy^212 = 1$$ What is the maximum perimeter of the rectangle?
I don't even know if I'm taking the right approach. So far, I've been trying to solve for $y$, giving me $y = sqrt12-(3/5)x$, and plugging that into the equation $P = 4x + 4y$, which should be the equation for the perimeter of an inscribed rectangle. I then took the derivative of $P$ after plugging in the equation for $y$, giving me $$P' = 4 - frac12x5sqrt12-(3/5)x.$$ To find a maximum, I'd set the equation to zero right? Well, I don't know where to go from this step, since simplifying from here only seems to make it harder.
Any help would be much appreciated, even a nudge in the right direction. I have no idea where to go from here, or even if I got to the right place. Thanks for your time
calculus optimization conic-sections rectangles
edited Nov 23 '15 at 11:09
Martin Sleziak
43.5k6113259
asked Nov 23 '15 at 3:31
Adam
2112
add a comment |Â
up vote
4
down vote
favorite
The problem I've been stuck on is this:
A rectangle is inscribed in the ellipse $$fracx^220 + fracy^212 = 1$$ What is the maximum perimeter of the rectangle?
I don't even know if I'm taking the right approach. So far, I've been trying to solve for $y$, giving me $y = sqrt12-(3/5)x$, and plugging that into the equation $P = 4x + 4y$, which should be the equation for the perimeter of an inscribed rectangle. I then took the derivative of $P$ after plugging in the equation for $y$, giving me $$P' = 4 - frac12x5sqrt12-(3/5)x.$$ To find a maximum, I'd set the equation to zero right? Well, I don't know where to go from this step, since simplifying from here only seems to make it harder.
Any help would be much appreciated, even a nudge in the right direction. I have no idea where to go from here, or even if I got to the right place. Thanks for your time
calculus optimization conic-sections rectangles
edited Nov 23 '15 at 11:09
Martin Sleziak
43.5k6113259
asked Nov 23 '15 at 3:31
Adam
2112
2
The general idea should work, though it is a little messy. You do not seem to have solved for $y$ in terms of $x$ correctly. Note that $y^2=12left(1-fracx^220right)$.
– André Nicolas
Nov 23 '15 at 3:38
Well, assuming your equation for $P'$ is correct, setting $P'=0$, moving the ugly fraction to one side, multiplying through by the denominator, and squaring both sides of the equation gives you a quadratic in $x$. And you know how to solve quadratics! EDIT: Just saw @AndréNicolas's comment - I think you should still get a quadratic?
– Zubin Mukerjee
Nov 23 '15 at 3:40
1
@ZubinMukerjee: Not only a quadratic, but one with no "$x$" term.
– André Nicolas
Nov 23 '15 at 4:12
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
The problem I've been stuck on is this:
A rectangle is inscribed in the ellipse $$fracx^220 + fracy^212 = 1$$ What is the maximum perimeter of the rectangle?
I don't even know if I'm taking the right approach. So far, I've been trying to solve for $y$, giving me $y = sqrt12-(3/5)x$, and plugging that into the equation $P = 4x + 4y$, which should be the equation for the perimeter of an inscribed rectangle. I then took the derivative of $P$ after plugging in the equation for $y$, giving me $$P' = 4 - frac12x5sqrt12-(3/5)x.$$ To find a maximum, I'd set the equation to zero right? Well, I don't know where to go from this step, since simplifying from here only seems to make it harder.
Any help would be much appreciated, even a nudge in the right direction. I have no idea where to go from here, or even if I got to the right place. Thanks for your time
calculus optimization conic-sections rectangles
edited Nov 23 '15 at 11:09
Martin Sleziak
43.5k6113259
asked Nov 23 '15 at 3:31
Adam
2112
The problem I've been stuck on is this:
A rectangle is inscribed in the ellipse $$fracx^220 + fracy^212 = 1$$ What is the maximum perimeter of the rectangle?
I don't even know if I'm taking the right approach. So far, I've been trying to solve for $y$, giving me $y = sqrt12-(3/5)x$, and plugging that into the equation $P = 4x + 4y$, which should be the equation for the perimeter of an inscribed rectangle. I then took the derivative of $P$ after plugging in the equation for $y$, giving me $$P' = 4 - frac12x5sqrt12-(3/5)x.$$ To find a maximum, I'd set the equation to zero right? Well, I don't know where to go from this step, since simplifying from here only seems to make it harder.
Any help would be much appreciated, even a nudge in the right direction. I have no idea where to go from here, or even if I got to the right place. Thanks for your time
calculus optimization conic-sections rectangles
edited Nov 23 '15 at 11:09
Martin Sleziak
43.5k6113259
asked Nov 23 '15 at 3:31
Adam
2112
edited Nov 23 '15 at 11:09
Martin Sleziak
43.5k6113259
edited Nov 23 '15 at 11:09
Martin Sleziak
43.5k6113259
edited Nov 23 '15 at 11:09
Martin Sleziak
43.5k6113259
43.5k6113259
asked Nov 23 '15 at 3:31
Adam
2112
asked Nov 23 '15 at 3:31
Adam
2112
asked Nov 23 '15 at 3:31
Adam
2112
2112
2
The general idea should work, though it is a little messy. You do not seem to have solved for $y$ in terms of $x$ correctly. Note that $y^2=12left(1-fracx^220right)$.
– André Nicolas
Nov 23 '15 at 3:38
Well, assuming your equation for $P'$ is correct, setting $P'=0$, moving the ugly fraction to one side, multiplying through by the denominator, and squaring both sides of the equation gives you a quadratic in $x$. And you know how to solve quadratics! EDIT: Just saw @AndréNicolas's comment - I think you should still get a quadratic?
– Zubin Mukerjee
Nov 23 '15 at 3:40
1
@ZubinMukerjee: Not only a quadratic, but one with no "$x$" term.
– André Nicolas
Nov 23 '15 at 4:12
add a comment |Â
2
The general idea should work, though it is a little messy. You do not seem to have solved for $y$ in terms of $x$ correctly. Note that $y^2=12left(1-fracx^220right)$.
– André Nicolas
Nov 23 '15 at 3:38
Well, assuming your equation for $P'$ is correct, setting $P'=0$, moving the ugly fraction to one side, multiplying through by the denominator, and squaring both sides of the equation gives you a quadratic in $x$. And you know how to solve quadratics! EDIT: Just saw @AndréNicolas's comment - I think you should still get a quadratic?
– Zubin Mukerjee
Nov 23 '15 at 3:40
1
@ZubinMukerjee: Not only a quadratic, but one with no "$x$" term.
– André Nicolas
Nov 23 '15 at 4:12
2
2
The general idea should work, though it is a little messy. You do not seem to have solved for $y$ in terms of $x$ correctly. Note that $y^2=12left(1-fracx^220right)$.
– André Nicolas
Nov 23 '15 at 3:38
The general idea should work, though it is a little messy. You do not seem to have solved for $y$ in terms of $x$ correctly. Note that $y^2=12left(1-fracx^220right)$.
– André Nicolas
Nov 23 '15 at 3:38
Well, assuming your equation for $P'$ is correct, setting $P'=0$, moving the ugly fraction to one side, multiplying through by the denominator, and squaring both sides of the equation gives you a quadratic in $x$. And you know how to solve quadratics! EDIT: Just saw @AndréNicolas's comment - I think you should still get a quadratic?
– Zubin Mukerjee
Nov 23 '15 at 3:40
Well, assuming your equation for $P'$ is correct, setting $P'=0$, moving the ugly fraction to one side, multiplying through by the denominator, and squaring both sides of the equation gives you a quadratic in $x$. And you know how to solve quadratics! EDIT: Just saw @AndréNicolas's comment - I think you should still get a quadratic?
– Zubin Mukerjee
Nov 23 '15 at 3:40
1
1
@ZubinMukerjee: Not only a quadratic, but one with no "$x$" term.
– André Nicolas
Nov 23 '15 at 4:12
@ZubinMukerjee: Not only a quadratic, but one with no "$x$" term.
– André Nicolas
Nov 23 '15 at 4:12
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
6
down vote
One simple way of solving this problem is by Lagrange multipliers method. Note that if $(x,y)$ is in the first quadrant on the ellipse $x^2/a^2+y^2/b^2 = 1$, then the perimeter of the inscribed rectangle represented by $(x,y)$ is simply $4(x+y)$. Therefore you want to maximize $x+y$ given the constraint that $x^2/a^2+y^2/b^2 = 1$. Define
$$
f(x,y,lambda) = x+y -lambdaleft(fracx^2a^2+fracy^2b^2-1right)
$$
Hence by maximizing $f$
$$
1 = frac2xlambdaa^2=frac2ylambdab^2Longrightarrow fracxa = fracybleft(fracabright)
$$
but then
$$1=fracx^2a^2+fracy^2b^2=fracy^2b^2left(1+fraca^2b^2right)Longrightarrow y=fracb^2sqrta^2+b^2, quad x=fraca^2sqrta^2+b^2$$
The maximum perimeter is therefore $4(x+y) = 4sqrta^2+b^2$.
answered Nov 23 '15 at 3:56
Hamed
4,333421
add a comment |Â
up vote
4
down vote
Let equation of ellipse be $displaystyle fracx^2a^2+fracy^2b^2 = 1;,$ Then we will take variable point $P,Q,R,S$
on that ellipse, and parametric Coordinate of Point $P(acos theta,bsin theta).$
Similarly $Q(-a cos theta,bsin theta)$ and $R(-a cos theta,-bsin theta)$ and $S(a cos theta,-bsin theta)$
So Paramteter of Recatangle is $$displaystyle P=4acos theta+4bsin theta =4(acos theta+bsin theta)leq 4sqrta^2+b^2.$$
Above we have used the formula $$bullet -sqrta^2+b^2leq (acos theta+bsintheta )leq sqrta^2+b^2$$
answered Nov 23 '15 at 4:02
juantheron
33.3k942122
add a comment |Â
up vote
1
down vote
Let me squeeze the ellipse into a circle:
$$fracx^212+fracy^212=1$$
And I would claim that the maximum perimeter rectangle inside the circle is the square. Its perimeter is
$$4sqrt2space r = 4sqrt2cdot2sqrt3 = 8sqrt6$$
Now let me recover the circle back to an ellipse. And the square is also stretched into a rectangle and one of its side is magnified by factor of $sqrt20/12=sqrt5/3$. And then the new perimeter is
$$8sqrt6cdotfracsqrt5sqrt3=8sqrt10$$
edited Nov 23 '15 at 4:08
Zubin Mukerjee
14k32456
answered Nov 23 '15 at 3:45
Kay K.
6,7211235
Why is it true that the rectangle that maximizes perimeter in the circle corresponds (by stretching in the $x$-direction) to the rectangle that maximizes perimeter in the ellipse? It's not obvious to me ...
– Zubin Mukerjee
Nov 23 '15 at 3:57
Well, it was true for area, but it doesn't seem to be true for perimeter based on Hamed's solution. Sorry about that.
– Kay K.
Nov 23 '15 at 4:01
Is it true for area?
– Zubin Mukerjee
Nov 23 '15 at 4:01
Yes. Otherwise, a square would have not been the maximum area inscribing rectangle inside a circle.
– Kay K.
Nov 23 '15 at 4:03
1
The stretching-into-a-circle is a cool way to do problems but you definitely have to be careful to note exactly why it is okay for you to reduce the problem to the circle version :)
– Zubin Mukerjee
Nov 23 '15 at 4:07
 |Â
show 3 more comments
up vote
0
down vote
All rectangles $[-a,a]times[-b,b]$ with given perimeter $p$ have the vertex $P=(a,b)$ on the line $$ell_p:quad a+b=pover4$$ of slope $-1$. Increasing $p$ means that $ell_p$ is translated north-east. The largest $p$ that can be realized for a $P$ on the given ellipse $$E:qquad f(x,y):=3x^2+5y^2=60tag1$$ is when $ell_p$ is tangent to $E$. We therefore have to find the point on $E$ in the first quadrant where
$nabla f(x,y)=(6x,10y)$ points due north-east. This enforces $y=3over5x$, so that we obtain from $(1)$ the point $P=1oversqrt2(5,3)$, leading to the maximal perimeter $$p_max=16sqrt2 .$$
answered Nov 23 '15 at 9:55
Christian Blatter
163k7107306
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
One simple way of solving this problem is by Lagrange multipliers method. Note that if $(x,y)$ is in the first quadrant on the ellipse $x^2/a^2+y^2/b^2 = 1$, then the perimeter of the inscribed rectangle represented by $(x,y)$ is simply $4(x+y)$. Therefore you want to maximize $x+y$ given the constraint that $x^2/a^2+y^2/b^2 = 1$. Define
$$
f(x,y,lambda) = x+y -lambdaleft(fracx^2a^2+fracy^2b^2-1right)
$$
Hence by maximizing $f$
$$
1 = frac2xlambdaa^2=frac2ylambdab^2Longrightarrow fracxa = fracybleft(fracabright)
$$
but then
$$1=fracx^2a^2+fracy^2b^2=fracy^2b^2left(1+fraca^2b^2right)Longrightarrow y=fracb^2sqrta^2+b^2, quad x=fraca^2sqrta^2+b^2$$
The maximum perimeter is therefore $4(x+y) = 4sqrta^2+b^2$.
answered Nov 23 '15 at 3:56
Hamed
4,333421
add a comment |Â
up vote
6
down vote
One simple way of solving this problem is by Lagrange multipliers method. Note that if $(x,y)$ is in the first quadrant on the ellipse $x^2/a^2+y^2/b^2 = 1$, then the perimeter of the inscribed rectangle represented by $(x,y)$ is simply $4(x+y)$. Therefore you want to maximize $x+y$ given the constraint that $x^2/a^2+y^2/b^2 = 1$. Define
$$
f(x,y,lambda) = x+y -lambdaleft(fracx^2a^2+fracy^2b^2-1right)
$$
Hence by maximizing $f$
$$
1 = frac2xlambdaa^2=frac2ylambdab^2Longrightarrow fracxa = fracybleft(fracabright)
$$
but then
$$1=fracx^2a^2+fracy^2b^2=fracy^2b^2left(1+fraca^2b^2right)Longrightarrow y=fracb^2sqrta^2+b^2, quad x=fraca^2sqrta^2+b^2$$
The maximum perimeter is therefore $4(x+y) = 4sqrta^2+b^2$.
answered Nov 23 '15 at 3:56
Hamed
4,333421
add a comment |Â
up vote
6
down vote
up vote
6
down vote
One simple way of solving this problem is by Lagrange multipliers method. Note that if $(x,y)$ is in the first quadrant on the ellipse $x^2/a^2+y^2/b^2 = 1$, then the perimeter of the inscribed rectangle represented by $(x,y)$ is simply $4(x+y)$. Therefore you want to maximize $x+y$ given the constraint that $x^2/a^2+y^2/b^2 = 1$. Define
$$
f(x,y,lambda) = x+y -lambdaleft(fracx^2a^2+fracy^2b^2-1right)
$$
Hence by maximizing $f$
$$
1 = frac2xlambdaa^2=frac2ylambdab^2Longrightarrow fracxa = fracybleft(fracabright)
$$
but then
$$1=fracx^2a^2+fracy^2b^2=fracy^2b^2left(1+fraca^2b^2right)Longrightarrow y=fracb^2sqrta^2+b^2, quad x=fraca^2sqrta^2+b^2$$
The maximum perimeter is therefore $4(x+y) = 4sqrta^2+b^2$.
answered Nov 23 '15 at 3:56
Hamed
4,333421
One simple way of solving this problem is by Lagrange multipliers method. Note that if $(x,y)$ is in the first quadrant on the ellipse $x^2/a^2+y^2/b^2 = 1$, then the perimeter of the inscribed rectangle represented by $(x,y)$ is simply $4(x+y)$. Therefore you want to maximize $x+y$ given the constraint that $x^2/a^2+y^2/b^2 = 1$. Define
$$
f(x,y,lambda) = x+y -lambdaleft(fracx^2a^2+fracy^2b^2-1right)
$$
Hence by maximizing $f$
$$
1 = frac2xlambdaa^2=frac2ylambdab^2Longrightarrow fracxa = fracybleft(fracabright)
$$
but then
$$1=fracx^2a^2+fracy^2b^2=fracy^2b^2left(1+fraca^2b^2right)Longrightarrow y=fracb^2sqrta^2+b^2, quad x=fraca^2sqrta^2+b^2$$
The maximum perimeter is therefore $4(x+y) = 4sqrta^2+b^2$.
answered Nov 23 '15 at 3:56
Hamed
4,333421
answered Nov 23 '15 at 3:56
Hamed
4,333421
answered Nov 23 '15 at 3:56
Hamed
4,333421
answered Nov 23 '15 at 3:56
Hamed
4,333421
4,333421
add a comment |Â
add a comment |Â
up vote
4
down vote
Let equation of ellipse be $displaystyle fracx^2a^2+fracy^2b^2 = 1;,$ Then we will take variable point $P,Q,R,S$
on that ellipse, and parametric Coordinate of Point $P(acos theta,bsin theta).$
Similarly $Q(-a cos theta,bsin theta)$ and $R(-a cos theta,-bsin theta)$ and $S(a cos theta,-bsin theta)$
So Paramteter of Recatangle is $$displaystyle P=4acos theta+4bsin theta =4(acos theta+bsin theta)leq 4sqrta^2+b^2.$$
Above we have used the formula $$bullet -sqrta^2+b^2leq (acos theta+bsintheta )leq sqrta^2+b^2$$
answered Nov 23 '15 at 4:02
juantheron
33.3k942122
add a comment |Â
up vote
4
down vote
Let equation of ellipse be $displaystyle fracx^2a^2+fracy^2b^2 = 1;,$ Then we will take variable point $P,Q,R,S$
on that ellipse, and parametric Coordinate of Point $P(acos theta,bsin theta).$
Similarly $Q(-a cos theta,bsin theta)$ and $R(-a cos theta,-bsin theta)$ and $S(a cos theta,-bsin theta)$
So Paramteter of Recatangle is $$displaystyle P=4acos theta+4bsin theta =4(acos theta+bsin theta)leq 4sqrta^2+b^2.$$
Above we have used the formula $$bullet -sqrta^2+b^2leq (acos theta+bsintheta )leq sqrta^2+b^2$$
answered Nov 23 '15 at 4:02
juantheron
33.3k942122
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Let equation of ellipse be $displaystyle fracx^2a^2+fracy^2b^2 = 1;,$ Then we will take variable point $P,Q,R,S$
on that ellipse, and parametric Coordinate of Point $P(acos theta,bsin theta).$
Similarly $Q(-a cos theta,bsin theta)$ and $R(-a cos theta,-bsin theta)$ and $S(a cos theta,-bsin theta)$
So Paramteter of Recatangle is $$displaystyle P=4acos theta+4bsin theta =4(acos theta+bsin theta)leq 4sqrta^2+b^2.$$
Above we have used the formula $$bullet -sqrta^2+b^2leq (acos theta+bsintheta )leq sqrta^2+b^2$$
answered Nov 23 '15 at 4:02
juantheron
33.3k942122
Let equation of ellipse be $displaystyle fracx^2a^2+fracy^2b^2 = 1;,$ Then we will take variable point $P,Q,R,S$
on that ellipse, and parametric Coordinate of Point $P(acos theta,bsin theta).$
Similarly $Q(-a cos theta,bsin theta)$ and $R(-a cos theta,-bsin theta)$ and $S(a cos theta,-bsin theta)$
So Paramteter of Recatangle is $$displaystyle P=4acos theta+4bsin theta =4(acos theta+bsin theta)leq 4sqrta^2+b^2.$$
Above we have used the formula $$bullet -sqrta^2+b^2leq (acos theta+bsintheta )leq sqrta^2+b^2$$
answered Nov 23 '15 at 4:02
juantheron
33.3k942122
answered Nov 23 '15 at 4:02
juantheron
33.3k942122
answered Nov 23 '15 at 4:02
juantheron
33.3k942122
answered Nov 23 '15 at 4:02
juantheron
33.3k942122
33.3k942122
add a comment |Â
add a comment |Â
up vote
1
down vote
Let me squeeze the ellipse into a circle:
$$fracx^212+fracy^212=1$$
And I would claim that the maximum perimeter rectangle inside the circle is the square. Its perimeter is
$$4sqrt2space r = 4sqrt2cdot2sqrt3 = 8sqrt6$$
Now let me recover the circle back to an ellipse. And the square is also stretched into a rectangle and one of its side is magnified by factor of $sqrt20/12=sqrt5/3$. And then the new perimeter is
$$8sqrt6cdotfracsqrt5sqrt3=8sqrt10$$
edited Nov 23 '15 at 4:08
Zubin Mukerjee
14k32456
answered Nov 23 '15 at 3:45
Kay K.
6,7211235
Why is it true that the rectangle that maximizes perimeter in the circle corresponds (by stretching in the $x$-direction) to the rectangle that maximizes perimeter in the ellipse? It's not obvious to me ...
– Zubin Mukerjee
Nov 23 '15 at 3:57
Well, it was true for area, but it doesn't seem to be true for perimeter based on Hamed's solution. Sorry about that.
– Kay K.
Nov 23 '15 at 4:01
Is it true for area?
– Zubin Mukerjee
Nov 23 '15 at 4:01
Yes. Otherwise, a square would have not been the maximum area inscribing rectangle inside a circle.
– Kay K.
Nov 23 '15 at 4:03
1
The stretching-into-a-circle is a cool way to do problems but you definitely have to be careful to note exactly why it is okay for you to reduce the problem to the circle version :)
– Zubin Mukerjee
Nov 23 '15 at 4:07
 |Â
show 3 more comments
up vote
1
down vote
Let me squeeze the ellipse into a circle:
$$fracx^212+fracy^212=1$$
And I would claim that the maximum perimeter rectangle inside the circle is the square. Its perimeter is
$$4sqrt2space r = 4sqrt2cdot2sqrt3 = 8sqrt6$$
Now let me recover the circle back to an ellipse. And the square is also stretched into a rectangle and one of its side is magnified by factor of $sqrt20/12=sqrt5/3$. And then the new perimeter is
$$8sqrt6cdotfracsqrt5sqrt3=8sqrt10$$
edited Nov 23 '15 at 4:08
Zubin Mukerjee
14k32456
answered Nov 23 '15 at 3:45
Kay K.
6,7211235
Why is it true that the rectangle that maximizes perimeter in the circle corresponds (by stretching in the $x$-direction) to the rectangle that maximizes perimeter in the ellipse? It's not obvious to me ...
– Zubin Mukerjee
Nov 23 '15 at 3:57
Well, it was true for area, but it doesn't seem to be true for perimeter based on Hamed's solution. Sorry about that.
– Kay K.
Nov 23 '15 at 4:01
Is it true for area?
– Zubin Mukerjee
Nov 23 '15 at 4:01
Yes. Otherwise, a square would have not been the maximum area inscribing rectangle inside a circle.
– Kay K.
Nov 23 '15 at 4:03
1
The stretching-into-a-circle is a cool way to do problems but you definitely have to be careful to note exactly why it is okay for you to reduce the problem to the circle version :)
– Zubin Mukerjee
Nov 23 '15 at 4:07
 |Â
show 3 more comments
up vote
1
down vote
up vote
1
down vote
Let me squeeze the ellipse into a circle:
$$fracx^212+fracy^212=1$$
And I would claim that the maximum perimeter rectangle inside the circle is the square. Its perimeter is
$$4sqrt2space r = 4sqrt2cdot2sqrt3 = 8sqrt6$$
Now let me recover the circle back to an ellipse. And the square is also stretched into a rectangle and one of its side is magnified by factor of $sqrt20/12=sqrt5/3$. And then the new perimeter is
$$8sqrt6cdotfracsqrt5sqrt3=8sqrt10$$
edited Nov 23 '15 at 4:08
Zubin Mukerjee
14k32456
answered Nov 23 '15 at 3:45
Kay K.
6,7211235
Let me squeeze the ellipse into a circle:
$$fracx^212+fracy^212=1$$
And I would claim that the maximum perimeter rectangle inside the circle is the square. Its perimeter is
$$4sqrt2space r = 4sqrt2cdot2sqrt3 = 8sqrt6$$
Now let me recover the circle back to an ellipse. And the square is also stretched into a rectangle and one of its side is magnified by factor of $sqrt20/12=sqrt5/3$. And then the new perimeter is
$$8sqrt6cdotfracsqrt5sqrt3=8sqrt10$$
edited Nov 23 '15 at 4:08
Zubin Mukerjee
14k32456
answered Nov 23 '15 at 3:45
Kay K.
6,7211235
edited Nov 23 '15 at 4:08
Zubin Mukerjee
14k32456
edited Nov 23 '15 at 4:08
Zubin Mukerjee
14k32456
edited Nov 23 '15 at 4:08
Zubin Mukerjee
14k32456
14k32456
answered Nov 23 '15 at 3:45
Kay K.
6,7211235
answered Nov 23 '15 at 3:45
Kay K.
6,7211235
answered Nov 23 '15 at 3:45
Kay K.
6,7211235
6,7211235
Why is it true that the rectangle that maximizes perimeter in the circle corresponds (by stretching in the $x$-direction) to the rectangle that maximizes perimeter in the ellipse? It's not obvious to me ...
– Zubin Mukerjee
Nov 23 '15 at 3:57
Well, it was true for area, but it doesn't seem to be true for perimeter based on Hamed's solution. Sorry about that.
– Kay K.
Nov 23 '15 at 4:01
Is it true for area?
– Zubin Mukerjee
Nov 23 '15 at 4:01
Yes. Otherwise, a square would have not been the maximum area inscribing rectangle inside a circle.
– Kay K.
Nov 23 '15 at 4:03
1
The stretching-into-a-circle is a cool way to do problems but you definitely have to be careful to note exactly why it is okay for you to reduce the problem to the circle version :)
– Zubin Mukerjee
Nov 23 '15 at 4:07
 |Â
show 3 more comments
Why is it true that the rectangle that maximizes perimeter in the circle corresponds (by stretching in the $x$-direction) to the rectangle that maximizes perimeter in the ellipse? It's not obvious to me ...
– Zubin Mukerjee
Nov 23 '15 at 3:57
Well, it was true for area, but it doesn't seem to be true for perimeter based on Hamed's solution. Sorry about that.
– Kay K.
Nov 23 '15 at 4:01
Is it true for area?
– Zubin Mukerjee
Nov 23 '15 at 4:01
Yes. Otherwise, a square would have not been the maximum area inscribing rectangle inside a circle.
– Kay K.
Nov 23 '15 at 4:03
1
The stretching-into-a-circle is a cool way to do problems but you definitely have to be careful to note exactly why it is okay for you to reduce the problem to the circle version :)
– Zubin Mukerjee
Nov 23 '15 at 4:07
Why is it true that the rectangle that maximizes perimeter in the circle corresponds (by stretching in the $x$-direction) to the rectangle that maximizes perimeter in the ellipse? It's not obvious to me ...
– Zubin Mukerjee
Nov 23 '15 at 3:57
Why is it true that the rectangle that maximizes perimeter in the circle corresponds (by stretching in the $x$-direction) to the rectangle that maximizes perimeter in the ellipse? It's not obvious to me ...
– Zubin Mukerjee
Nov 23 '15 at 3:57
Well, it was true for area, but it doesn't seem to be true for perimeter based on Hamed's solution. Sorry about that.
– Kay K.
Nov 23 '15 at 4:01
Well, it was true for area, but it doesn't seem to be true for perimeter based on Hamed's solution. Sorry about that.
– Kay K.
Nov 23 '15 at 4:01
Is it true for area?
– Zubin Mukerjee
Nov 23 '15 at 4:01
Is it true for area?
– Zubin Mukerjee
Nov 23 '15 at 4:01
Yes. Otherwise, a square would have not been the maximum area inscribing rectangle inside a circle.
– Kay K.
Nov 23 '15 at 4:03
Yes. Otherwise, a square would have not been the maximum area inscribing rectangle inside a circle.
– Kay K.
Nov 23 '15 at 4:03
1
1
The stretching-into-a-circle is a cool way to do problems but you definitely have to be careful to note exactly why it is okay for you to reduce the problem to the circle version :)
– Zubin Mukerjee
Nov 23 '15 at 4:07
The stretching-into-a-circle is a cool way to do problems but you definitely have to be careful to note exactly why it is okay for you to reduce the problem to the circle version :)
– Zubin Mukerjee
Nov 23 '15 at 4:07
 |Â
show 3 more comments
up vote
0
down vote
All rectangles $[-a,a]times[-b,b]$ with given perimeter $p$ have the vertex $P=(a,b)$ on the line $$ell_p:quad a+b=pover4$$ of slope $-1$. Increasing $p$ means that $ell_p$ is translated north-east. The largest $p$ that can be realized for a $P$ on the given ellipse $$E:qquad f(x,y):=3x^2+5y^2=60tag1$$ is when $ell_p$ is tangent to $E$. We therefore have to find the point on $E$ in the first quadrant where
$nabla f(x,y)=(6x,10y)$ points due north-east. This enforces $y=3over5x$, so that we obtain from $(1)$ the point $P=1oversqrt2(5,3)$, leading to the maximal perimeter $$p_max=16sqrt2 .$$
answered Nov 23 '15 at 9:55
Christian Blatter
163k7107306
add a comment |Â
up vote
0
down vote
All rectangles $[-a,a]times[-b,b]$ with given perimeter $p$ have the vertex $P=(a,b)$ on the line $$ell_p:quad a+b=pover4$$ of slope $-1$. Increasing $p$ means that $ell_p$ is translated north-east. The largest $p$ that can be realized for a $P$ on the given ellipse $$E:qquad f(x,y):=3x^2+5y^2=60tag1$$ is when $ell_p$ is tangent to $E$. We therefore have to find the point on $E$ in the first quadrant where
$nabla f(x,y)=(6x,10y)$ points due north-east. This enforces $y=3over5x$, so that we obtain from $(1)$ the point $P=1oversqrt2(5,3)$, leading to the maximal perimeter $$p_max=16sqrt2 .$$
answered Nov 23 '15 at 9:55
Christian Blatter
163k7107306
add a comment |Â
up vote
0
down vote
up vote
0
down vote
All rectangles $[-a,a]times[-b,b]$ with given perimeter $p$ have the vertex $P=(a,b)$ on the line $$ell_p:quad a+b=pover4$$ of slope $-1$. Increasing $p$ means that $ell_p$ is translated north-east. The largest $p$ that can be realized for a $P$ on the given ellipse $$E:qquad f(x,y):=3x^2+5y^2=60tag1$$ is when $ell_p$ is tangent to $E$. We therefore have to find the point on $E$ in the first quadrant where
$nabla f(x,y)=(6x,10y)$ points due north-east. This enforces $y=3over5x$, so that we obtain from $(1)$ the point $P=1oversqrt2(5,3)$, leading to the maximal perimeter $$p_max=16sqrt2 .$$
answered Nov 23 '15 at 9:55
Christian Blatter
163k7107306
All rectangles $[-a,a]times[-b,b]$ with given perimeter $p$ have the vertex $P=(a,b)$ on the line $$ell_p:quad a+b=pover4$$ of slope $-1$. Increasing $p$ means that $ell_p$ is translated north-east. The largest $p$ that can be realized for a $P$ on the given ellipse $$E:qquad f(x,y):=3x^2+5y^2=60tag1$$ is when $ell_p$ is tangent to $E$. We therefore have to find the point on $E$ in the first quadrant where
$nabla f(x,y)=(6x,10y)$ points due north-east. This enforces $y=3over5x$, so that we obtain from $(1)$ the point $P=1oversqrt2(5,3)$, leading to the maximal perimeter $$p_max=16sqrt2 .$$
answered Nov 23 '15 at 9:55
Christian Blatter
163k7107306
answered Nov 23 '15 at 9:55
Christian Blatter
163k7107306
answered Nov 23 '15 at 9:55
Christian Blatter
163k7107306
answered Nov 23 '15 at 9:55
Christian Blatter
163k7107306
163k7107306
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1542234%2fhow-do-i-find-the-maximum-perimeter-of-a-rectangle-inscribed-in-an-ellipse%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
The general idea should work, though it is a little messy. You do not seem to have solved for $y$ in terms of $x$ correctly. Note that $y^2=12left(1-fracx^220right)$.
– André Nicolas
Nov 23 '15 at 3:38
Well, assuming your equation for $P'$ is correct, setting $P'=0$, moving the ugly fraction to one side, multiplying through by the denominator, and squaring both sides of the equation gives you a quadratic in $x$. And you know how to solve quadratics! EDIT: Just saw @AndréNicolas's comment - I think you should still get a quadratic?
– Zubin Mukerjee
Nov 23 '15 at 3:40
1
@ZubinMukerjee: Not only a quadratic, but one with no "$x$" term.
– André Nicolas
Nov 23 '15 at 4:12