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MarcLeclair
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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?
 one year ago
 one year ago
MarcLeclair Group Title
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?
 one year ago
 one year ago

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MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix
 one year ago

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

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
or wait forget that last equation :P
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1355978532328:dwLet's see if we can get the box drawn accurately.
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
yeah I have it drawn but im still wondering if its symmetrical :P
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
dw:1355978702013:dwUnderstand how I labeled those lengths?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Yes 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.
 one year ago

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

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
It'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.
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
yeah 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
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Yah :) 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.
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
yeah 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?
 one year ago

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

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
so the constraint is already labelled as y=5x^2! smart :P
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Ah yes, that's convenient :)
 one year ago

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

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

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
so 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?
 one year ago

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

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
my answer is still wrong x.x and yes it does , hahah
 one year ago

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

zepdrix Group TitleBest 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
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
thanks a lot youre REALLLY helpful, i might pass my final thanks to you haha
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
We 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. :)
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So now what we want to do is ummm
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Plug our critical point into our constraint, to find a corresponding length for Y.
 one year ago

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

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Bah yer too fast! lol
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
So 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.
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
Which i found as 20/6
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
:) thanks a lot mate!!!
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
Yay team! :) No problem.
 one year ago

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

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
still 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!
 one year ago

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

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