- anonymous

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?

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

- anonymous

I can't seem to find the the constraint and the good equation to make the problem work :/ except for x=-y - y

- anonymous

or wait forget that last equation :P

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- zepdrix

|dw:1355978532328:dw|Let's see if we can get the box drawn accurately.

- anonymous

yeah I have it drawn but im still wondering if its symmetrical :P

- zepdrix

|dw:1355978702013:dw|Understand how I labeled those lengths?

- zepdrix

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.

- anonymous

yeah it make sense, but shouldnt it be 1/2 x ? or its just 2 rectangles then?

- zepdrix

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.

- anonymous

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

- zepdrix

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.

- anonymous

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?

- zepdrix

The constraint is the parabola, we are constrained to the shape of the parabola. Otherwise the box could just grow and grow and grow.

- anonymous

so the constraint is already labelled as y=5-x^2! smart :P

- zepdrix

Ah yes, that's convenient :)

- anonymous

so then, if that is my constraint a= 2x(y) and I can replace y with 5-x^2?

- zepdrix

\[\huge A=2x\color{orangered}{y}\]Yes very good :)

- anonymous

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?

- zepdrix

Woops, I think your 10x will turn into 10 when you differentiate it, right?

- anonymous

my answer is still wrong x.x and yes it does , hahah

- anonymous

nvm my answer is 2sqrt(10/6), but I have to multiply by 2 right?

- zepdrix

\[\large A=10x-2x^3, \qquad A'=10-6x^2\]\[\large x=\pm \sqrt{\frac{5}{3}}\]So you're getting a critical point here, yes? Simplify your fraction :D heh

- anonymous

thanks a lot youre REALLLY helpful, i might pass my final thanks to you haha

- zepdrix

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. :)

- zepdrix

So now what we want to do is ummm

- zepdrix

Plug our critical point into our constraint, to find a corresponding length for Y.

- anonymous

we multiply by 2 because one x is only half the width, than plug it in y = 5-x^2 and were done :D

- zepdrix

Bah yer too fast! lol

- zepdrix

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.

- anonymous

Which i found as 20/6

- anonymous

:) thanks a lot mate!!!

- zepdrix

Yay team! :) No problem.

- zepdrix

But again, simplify that ugly fraction! lol
20/6 = 10/3

- anonymous

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!

- zepdrix

\[\large Length=2\sqrt{\frac{5}{3}}, \qquad Width=\frac{10}{3}\]

- zepdrix

Yah related rates are tough to get a grip on.
They show up in many many different forms. :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.