Optimization under differentiation

[don1231915]

Optimization under differentiation!

Homework Statement

OK
I have a upside down looking curve structure (½ ellipse). It has the following specifications:
The building has a rectangular base 150m long and 72m wide. The max height of the structure should not exceed 75% of its width or be less than half the width. And the min. height of a room in the building is 2.5m

Homework Equations

The first objective was to create a model when the height is 36m
I did it using the equation of the ellipse and my graph looks like: http://imageupload.org/pt-112919260786.html
The equation of this ellipse is y= sqrt(1296-x^2)

The QUESTION IS:

How do I find the maximum volume of the cuboid which would fit inside this curve?

The Attempt at a Solution

The only thing I have figured is
that we know nothing about the cuboid
So I assumed the width and height to be x and 2y
So the V = 300xy (x2 as the width is 2y)
and then I plugged in the equation of the ellipse into this equation, is that the right thing to do?
But then do I differentiate?? to find what? x?

PLEASE PLEASE HELP!

THANK YOU SO MUCH GUYS!

 

[HallsofIvy]

If you have [itex]y= b\sqrt{1- x^2/a^2}[/itex], half of an ellipse, then the height of any rectangle with base of length L, just fitting inside that ellipse, is [itex]h= b\sqrt{1- L^2/4a^2}[/itex] and the area is [itex]A= 2Lb\sqrt{2- L^2/4a^2}[/itex]. Differentiate that with respect to L to find thebase that will give maximum area.

 

[don1231915]

If you have [itex]y= b\sqrt{1- x^2/a^2}[/itex], half of an ellipse, then the height of any rectangle with base of length L, just fitting inside that ellipse, is [itex]h= b\sqrt{1- L^2/4a^2}[/itex] and the area is [itex]A= 2Lb\sqrt{2- L^2/4a^2}[/itex]. Differentiate that with respect to L to find thebase that will give maximum area.

Are a and b both 36? Also, I am having a really hard time differentiating it with respect to L.
Wondering if you can elaborate?? AND isn't it also supposed to be a cuboid??

Thank you so much for helping!