- #1

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- Homework Statement
- Find the volume of the largest rectangular parallelpiped that can be inscribed in the ellipsoid

##\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1##

- Relevant Equations
- No equation

Last edited:

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- Thread starter WMDhamnekar
- Start date

In summary: I think I see the problem. You must have thought he was saying ##1/3## was the maximum value of the function. No, it's the value of the function at its maximum value.In summary, the conversation discusses various mathematical approaches to solving a problem involving maximizing the volume of a rectangular parallelpiped inscribed in an ellipsoid. The use of substitution and Lagrange multipliers is suggested, with the suggestion to simplify the problem by setting certain variables to 1. The correct answer is found to be ##\frac{8abc}{3\sqrt{3}}##, but the main focus of the conversation is on understanding the method rather than the answer itself. The importance of understanding the constraints and setting up

- #1

- 376

- 28

- Homework Statement
- Find the volume of the largest rectangular parallelpiped that can be inscribed in the ellipsoid

##\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1##

- Relevant Equations
- No equation

Last edited:

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- #2

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Your problem has a multitude of short-cut possibilities. You distract yourself with your substitution.

With three equations in four unknowns you need a fourth. Any suggestions ? (hint: what is it you want to maximize ?)

General advice: if you don't succeed at first, try a simpler case. In this case you could set ##a=b=1## and maximize ##f=xy## subject to ##g = \sqrt{x^2+y^2}\le 1## with a well known answer -- but the point is to understand the method.

##\ ##

- #3

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Author provided the answer: ## \frac{8abc}{3\sqrt{3}}## I don't understand how was that computed?BvU said:

Your problem has a multitude of short-cut possibilities. You distract yourself with your substitution.

With three equations in four unknowns you need a fourth. Any suggestions ? (hint: what is it you want to maximize ?)

General advice: if you don't succeed at first, try a simpler case. In this case you could set ##a=b=1## and maximize ##f=xy## subject to ##g = \sqrt{x^2+y^2}\le 1## with a well known answer -- but the point is to understand the method.

##\ ##

- #4

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The exercise is intended to help you understand the method.

You chose to ignore my advice

(I could easily give you straight directions to get the answer, but, again, that's not the point of this exercise...)

To clarify:

useBvU said:hint: what is it you want to maximize

PS why did you post a picture instead of ##\LaTeX## typesetting ?

##\ ##

- #5

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Maximize f(x,y,z) = volume of rectangular parallelpiped.BvU said:The answer is NOT interesting (but you can check it with ##a=b=c=1## if you insist )

The exercise is intended to help you understand the method.

You chose to ignore my adviceandmy hint. Any particular reason ?

(I could easily give you straight directions to get the answer, but, again, that's not the point of this exercise...)

To clarify:

use View attachment 301199 to write ##f## as a function of ##x,y,z,\lambda##. Turns out to be very simple...

PS why did you post a picture instead of ##\LaTeX## typesetting ?

##\ ##

If we substitute a=b=c=1 in g(x,y,z), we get equation of sphere ##x^2 + y^2 + z^2 =1 \Rightarrow x=y=z=\frac{\sqrt{3}}{3} f(x,y,z) = \frac{1}{3\sqrt{3}} ## but as per author's answer it must be ##\frac{8abc}{3\sqrt{3}}= \frac{8}{3\sqrt{3}}##

I posted this question on other math help websites also having latex software, to get different ideas of solving the same question from various different math professionals and then analyze them according to its convenience to me. That's why I preferred to post a picture rather than using latex which is time consuming exercise.

- #6

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What did

##\ ##

- #7

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So what's the status now ? Factor 8 clear ?

##\ ##

##\ ##

- #8

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This question was solved.BvU said:So what's the status now ? Factor 8 clear ?

##\ ##

## \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = \frac{4xyz}{\lambda}##

and

##\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} =1##

##\therefore \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} =\frac13##

- #9

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I get $$ \begin{align*}

xyz & = 2\lambda {x^2\over a^2 }\ ,\\

xyz & =2\lambda{ y^2\over b^2} \ ,\\

xyz & =2\lambda{ z^2\over c^2}\quad \end{align*}

$$so not your

##\quad\ \ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = \frac{4xyz}{\lambda}##

but ##{x^2\over a^2 }={y^2\over b^2 }={z^2\over c^2 }##

and ##\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = {2\over 3} \frac{xyz}{\lambda}##

?

https://www.whitman.edu/mathematics/calculus_online/section14.08.html

Example 14.8.1 is the example I was referring to (with ##a=b=c=1##).

- #10

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##\qquad\therefore \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} =\frac13##

is very far from the constraint !

- #11

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I am very sorry. There are mistakes in the my answer given in #8.BvU said:

##\qquad\therefore \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} =\frac13##

is very far from the constraint !

Please read ## \frac{x^2}{a^2} = \frac{y^2}{b^2} = \frac{z^2}{c^2} = \frac{4xyz}{\lambda}##

and

##\therefore \frac{x^2}{a^2} = \frac{y^2}{b^2} =\frac{z^2}{c^2}= \frac13##

Last edited:

- #12

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Where does the ##1\over 3## come from all of a sudden?

- #13

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Please use some algebra to find out thatBvU said:Where does the ##1\over 3## come from all of a sudden?

- #14

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I'm not trying to find out that. I am asking you something

- #15

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Hi,BvU said:I'm not trying to find out that. I am asking you something

Let P(x, y, z) be a point on the ellipsoid. The 8 vertices of the rectangular parallelpiped are ## P(\pm x, \pm y , \pm z).## Therefore maximize Volume(V) of the rectangular parallelpiped## f(x,y,z)=V = (2x\cdot 2y \cdot 2z)= 8xyz##

given

##g (x,y,z) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}=1##

Now, read my post #1 and #11, use some algebra you will get the answer ##\frac{8abc}{3\sqrt{3}}## as given by author. I got the answer which correctly matches with the author's answer.

That's it.

- #16

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Your problem statement said

the largest rectangular parallelpiped that can be inscribed in the ellipsoid...

Which is different from

WMDhamnekar said:given

$$g (x,y,z) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}\ \ = \ \ 1$$

At some point (imho) you explicitly need to make the step from ##g\le 1## to ##g=1##. Perhaps you did, but then I missed that.

(see example 4 here)

Other than that, you did fine. Even without following the suggestions in #2 ...

[edit]Well, apart also from setting up the wrong ##f## in

##\ ##

- #17

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BvU said:Where does the ##1\over 3## come from all of a sudden?

3 equal numbers add up to 1. Are you really questioning how he deduced that each number is 1/3?BvU said:I'm not trying to find out that. I am asking you something

- #18

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Inscribed means that all corners are on the surface. That's why g=1.BvU said:At some point (imho) you explicitly need to make the step from ##g\le 1## to ##g=1##. Perhaps you did, but then I missed that.

- #19

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I think post #15 gives the answer, as well correcting the initial mistake in the objective function.

- #20

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A problem symmetric in several variables should have a solution symmetric in those variables.

Unlike the other problems this one requires a slight tweak at the beginning to make it symmetric. Just change the variables into three others: ##X=ax, Y=by, Z=cZ##, then the problem becomes find the volume of the cube whose corners are points on the sphere ##X^2+Y^2 +Z^2=1##, the volume of the parallipiped is then ##abc## times that.

The side of the cube whose corners are on a sphere of radius 1 is 2/√ 3

so giving a cube of volume 8/3√3 and parallelipiped volume 8abc/3√3.

The maxima in the other problems pwere obtainable by making other variables equal to the variable ##x## and finding the ##x## that maximised that singl-variable function, very simple. Of course you need to know Lagrange to be able to deal with the general case, and the right answer was always found - just I thought they were rather here's a problem about a formula, here's a method, turn the handle, here's another formula which is the solution, but the somewhat intriguing geometry of the functions went unnoticed.

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