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MarcLeclair
 3 years ago
A rectangle is inscribed with its base on the xaxis and its upper corners on the parabola y=5x^2 . What are the dimensions of such a rectangle with the greatest possible area?
MarcLeclair
 3 years ago
A rectangle is inscribed with its base on the xaxis and its upper corners on the parabola y=5x^2 . What are the dimensions of such a rectangle with the greatest possible area?

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MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0I can't seem to find the the constraint and the good equation to make the problem work :/ except for x=y  y

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0or wait forget that last equation :P

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1355978532328:dwLet's see if we can get the box drawn accurately.

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0yeah I have it drawn but im still wondering if its symmetrical :P

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1355978702013:dwUnderstand how I labeled those lengths?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Yes it's symmetrical, both corners have to touch the parabola. And in order for the top to be a straight line across, the points have to touch at the same places across the parabola.

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0yeah it make sense, but shouldnt it be 1/2 x ? or its just 2 rectangles then?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1It's one large rectangle, touching one end of the parabola and the other end as well. We're basically choosing a point along the parabola. The coordinates that correspond to that point we call (x,y)dw:1355978868031:dwAnd due to symmetry, it'll touch at the same spot on the other side.

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0yeah it makes sense, the y should be the same on both side, just the roots will be different but the x is the same but on the negative side xD

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Yah :) So if we want a function for AREA. We need to establish the LENGTH and WIDTH of this rectangle. The Width is pretty clearly Y. What about the Length? Remember, when dealing with length, we don't care about the negative. We can't have a wooden board that is 2 inches long.

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0yeah so it would be 2x?. However I'll stop you for a sec, everytime I do an optimization i set myself a constraint but there's none here.. am i wrong?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1The constraint is the parabola, we are constrained to the shape of the parabola. Otherwise the box could just grow and grow and grow.

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0so the constraint is already labelled as y=5x^2! smart :P

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Ah yes, that's convenient :)

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0so then, if that is my constraint a= 2x(y) and I can replace y with 5x^2?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\huge A=2x\color{orangered}{y}\]Yes very good :)

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0so then i find my derivative as a'(x) = 6x^2 10x. I get my 0 as +/ sqrt(10/6), do I need to x2 for my full width?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Woops, I think your 10x will turn into 10 when you differentiate it, right?

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0my answer is still wrong x.x and yes it does , hahah

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0nvm my answer is 2sqrt(10/6), but I have to multiply by 2 right?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large A=10x2x^3, \qquad A'=106x^2\]\[\large x=\pm \sqrt{\frac{5}{3}}\]So you're getting a critical point here, yes? Simplify your fraction :D heh

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0thanks a lot youre REALLLY helpful, i might pass my final thanks to you haha

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1We only care about the positive square root (since it's a length). So I guess I didn't need to worry about the p/m sign. :)

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1So now what we want to do is ummm

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Plug our critical point into our constraint, to find a corresponding length for Y.

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0we multiply by 2 because one x is only half the width, than plug it in y = 5x^2 and were done :D

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1So in this problem, they don't want the MAXIMUM area, they just want the LENGTH and WIDTH of the box. We're already half done. We found x. Or rather as you pointed out, 2x. Now we just need Y.

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0Which i found as 20/6

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0:) thanks a lot mate!!!

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Yay team! :) No problem.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1But again, simplify that ugly fraction! lol 20/6 = 10/3

MarcLeclair
 3 years ago
Best ResponseYou've already chosen the best response.0still ugly in my world. I don't know how you do it but I can never see through this small number of information im given. Or I barely can hahaha. Thanks!

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large Length=2\sqrt{\frac{5}{3}}, \qquad Width=\frac{10}{3}\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.1Yah related rates are tough to get a grip on. They show up in many many different forms. :)
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