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

OCW Scholar - Single Variable Calculus
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I can't seem to find the the constraint and the good equation to make the problem work :/ except for x=-y - y
or wait forget that last equation :P

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Other answers:

|dw:1355978532328:dw|Let's see if we can get the box drawn accurately.
yeah I have it drawn but im still wondering if its symmetrical :P
|dw:1355978702013:dw|Understand how I labeled those lengths?
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.
yeah it make sense, but shouldnt it be 1/2 x ? or its just 2 rectangles then?
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:dw|And due to symmetry, it'll touch at the same spot on the other side.
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
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.
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?
The constraint is the parabola, we are constrained to the shape of the parabola. Otherwise the box could just grow and grow and grow.
so the constraint is already labelled as y=5-x^2! smart :P
Ah yes, that's convenient :)
so then, if that is my constraint a= 2x(y) and I can replace y with 5-x^2?
\[\huge A=2x\color{orangered}{y}\]Yes very good :)
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?
Woops, I think your 10x will turn into 10 when you differentiate it, right?
my answer is still wrong x.x and yes it does , hahah
nvm my answer is 2sqrt(10/6), but I have to multiply by 2 right?
\[\large A=10x-2x^3, \qquad A'=10-6x^2\]\[\large x=\pm \sqrt{\frac{5}{3}}\]So you're getting a critical point here, yes? Simplify your fraction :D heh
thanks a lot youre REALLLY helpful, i might pass my final thanks to you haha
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. :)
So now what we want to do is ummm
Plug our critical point into our constraint, to find a corresponding length for Y.
we multiply by 2 because one x is only half the width, than plug it in y = 5-x^2 and were done :D
Bah yer too fast! lol
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.
Which i found as 20/6
:) thanks a lot mate!!!
Yay team! :) No problem.
But again, simplify that ugly fraction! lol 20/6 = 10/3
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!
\[\large Length=2\sqrt{\frac{5}{3}}, \qquad Width=\frac{10}{3}\]
Yah related rates are tough to get a grip on. They show up in many many different forms. :)

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