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or wait forget that last equation :P

|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?

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

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 :)

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?

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

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

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

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

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