Optimization Problem: Largest Rectangle Inscribed in an Ellipse

Optimization Problem: Largest Rectangle Inscribed in an Ellipse
📐 Maximize Your Geometry Skills! 📐
In this video, we tackle a calculus optimization question: finding the largest rectangle that can be inscribed within the ellipse defined by the equation (x^2) / 4 + (y^2) / 9 = 1
What You’ll Learn:
Understanding the Ellipse: Get a clear overview of the ellipse and how it relates to the rectangle we want to inscribe.
Setting Up the Optimization Problem: Learn how to formulate the area of the rectangle in terms of the ellipse’s dimensions.
Applying Calculus Techniques: Follow along as I derive the area function, find critical points, and determine the dimensions of the rectangle that maximize the area.
Why Watch This Video?
Ideal for Students: Perfect for high school and college students looking to strengthen their understanding of optimization in calculus.
Clear Explanations: Enjoy step-by-step guidance that makes complex concepts easy to grasp.
Real-World Applications: Discover how these optimization principles can be applied in various fields, including engineering and design.
📈 Engage with the Content:
LIKE this video if it helps enhance your understanding of optimization!
SHARE with friends or classmates eager to learn more about calculus!
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#Optimization #Ellipse #Calculus #MaximizingArea #Mathematics #MathTutorial #EducationalContent #LearningCalculus #ProblemSolving #HighSchoolMath #CollegeCalculus #RealWorldApplications #Geometry

[@yanarisutandey3601]

Hell nah man. My professor gave a problem like this for an exam tonight except it was an ellipsoid and we had to maximize a rectangular box with in it's constraints. Talk about PISSED. Hard ass exam.

[@sumairsunny2801]

From equation of an ellipse: a=2, b =3: by observation the area of the rectangle inscribed inside the ellipse is 2*a*cos(theta) * 2*b*sin(theta).Which is 4*a*b*sin(theta)*cos(theta). since 4*a*b are constants of the ellipse we don't really need to care about them (as the ellipse changes as long as the area of the rectangle has a*b in its formula, the rectangle will also change accordingly). What will determine the biggest (or largest area) rectangle will be the biggest possible value of cos(theta)*sin(theta). Even if you're bad at math, just plug in values for theta and see what the biggest value of (sin(theta*cos(theta)) comes to. you will notice the largest value is 0.5 @ theta = 45 degrees... so largest AREA of rectangle inscribed in an ellipse = 4*a*b*0.5 = 2ab.

[@gidonkessler1799]

very cool

[@animeshbhargava2625]

Please make a video of cauchy-riemann equation Your videos are really helpful

[@cctang4922]

I think it is easier to consider the square of the area . After substituting y in terms of x in the equation of square of area, we can take the derivative of square of A with respect to x and set it to zero, the critical values of x can still be obtained . One of the advantages of doing this is that the differentiation is more simple. The other is that there is no need to discuss the numerator of the derivative as in your method.

[@Crissybooable]

hate the sharpie sound, love the learning

[@GlaceGFX]

I think you can just set the area function equal to another function but it is squared to remove the square root in order to make differentiation easier.

[@TAYLOR-113]

Dude, I have a quiz on optimization tomorrow. Perfect timing lol

[@patrickjmt]

My students have a test this week over it :) Good luck on your quiz!

[@leterpou7111]

+Taylor S oh noes i have test this week

[@DogeFrom2014]

+Taylor S Dude same, and one of the practice questions was this exact question.

[@gofio123456]

Amazing work you just made my studentlife much easier ! 👍

[@DruboOpticsAndwatch]

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[@DruboOpticsAndwatch]

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[@DruboOpticsAndwatch]

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[@logeshwarsaravanakumar6624]

Guys, take f as 4xy and g has given ellipse equation and use Lagrangian multiplier method i.e F = f + lambda g. It will be pretty ezy✌️

[@heliocentric1756]

A=4xy so A'=4y+4xy'=0 implies y'=-y/x (Where A' and y' are the derivatives with respect to [wrt] x) Differentiate the equation of the ellipse wrt x: (2x/4)+(2yy'/9)=0 , substitute -y/x for y': (x/2)-(2y^2/9x)=0, so 9x^2=4y^2 ...(1) The equation of the ellipse can be simplified to: 9x^2+4y^2=36 ..(2) Equations (1) and (2) give x=sqrt(2) and y=3/sqrt(2). After showing that these values of x and y actually yield a maximum area: A=12

[@annapabreu]

You have already helped a brazilian student! Many thanks

[@1111K-q5g]

Patrick JMT - that was wow! Your the best !

[@kindaamazing3796]

you have saved the life of a Pakistani student!thnx a lot

[@SADDAMHUSSAIN-mw3cv]

Very thank you respected sir...

[@imaginaryobserver]

You are amazing.

[@grigorystepanov5436]

I really enjoy ur content thanks

[@menousakis]

thank you for those videos

[@vatsalyataneja1408]

Thanks brother!

[@reddragon1790]

take the co ordinates as (2cos@, 3sin@) then the problem becomes a piece of cake.

[@colinsilver1041]

God I hate fractions. I try to get rid of them as soon as possible. I needed clarity on the "constraint" function, once I get pass that, I'm OK.

[@turkeyroast2223]

Isn't this easier with Lagrange Multipliers?

[@mwrkhan]

+Jaxkelington Jenkins Yes. Do you happen to be a student of economics by any chance?

[@turkeyroast2223]

nope, Lagrange Multipliers was in my calculus textbook

[@nathangillespie9704]

I believe that your original area equation is incorrect, I think it should be [ A=2x*2y OR A=2(xy)]. Correct me if I'm wrong though.

[@hermanlegaspi5713]

It's actually A=4xy.

[@angelan.9779]

super helpful, thank you so much

[@Bangalibabu95]

Can't we assume the co-ordinate as (2x, 2y) or (x/2, y/2) in place of (x,y) ?

[@cctang4922]

Correction, it should be denominator not numerator. I make a mistake.

[@hass_pkmkb]

Thanks

[@alibuxbaloch3645]

sir your emazzing realy i like ur all videos i am from pakistan

[@ashishbhardwaj2003]

Freedom to balochistan soon

[@roseb2105]

im confused when you took the derivative of x squared/4 if you use the quotient rule you get 1/16 and derivative of 1 is equal zero I get how you get half but which answer and method is it?

[@cvismenu]

Thank you

[@patrickjmt]

You're welcome

[@rgqwerty63]

Who says the rectangle has to be oriented like that? What if I decide its slanted?

[@ibrahimismael7497]

It will be the same, the drawing is just for explanation.

[@rgqwerty63]

+Ibrahim Ismael no it's not the same. if it were a circle it would be the same since you could rotate arbitrarily. but an ellipse has minor and major axis which he's assumed the rectangles edges are parallel to.

[@ibrahimismael7497]

I agree with you, but maybe it's because when you make the rectangle taller for ex. the y increases and x decreases, so it's like inverse relationship that wouldn't change the area.

[@ibrahimismael7497]

just point of view

[@SmokingBlackBear]

I need to do this in one quadrant with a triangle, Idk how

[@pablolopez3166]

Did you find out, mate? It's been like 2 years.

[@estleexin7584]

thanks

[@reydmore13100]

or simply 2ab

[@prestontao2]

lets get thumbs up all the way to 690 bc this is so useful

[@ibrahimismael7497]

You've just helped an Iraqi student :D

[@patrickjmt]

:) howdy from USA!

[@kiritokun5082]

need help with this problem, Find the dimension of rectangle of maximum Area A that can be inscribed in the portion of the parabola y*2 = 4px intercepted by the line x = a......pls email me with a detailed solution:)

[@mahdude7826]

Kill me i got an exam tomorrow at 7am

[@roseb2105]

the concept isnt to hard to get but the calculations are tedious and messy

[@heliocentric1756]

1:40 Nice face there :)

[@tejasshenoy2095]

Just hope priyanka doesn't get fooled lol

[@harikrishnametta8549]

wrong!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[@realhriday]

an avg indian math student can solve this in 2 mins by squaring the area and then taking the derivative and putting it equal to 0