q6.) i think you mean

We can take a shortcut in this problem by stating that if one of the co-ordinates of the biggest rectangle are (x,y) then there must be another one that is (-x,y). Hence the problem can be reduced to just by limiting ourselves to the first quadrant only and doubling our area because of symmetry about y axis. We will call this half Area

:




(yeah you could've used the product rule but I'm too lazy to use it)
}{\sqrt{9x^2-x^4}})
We require:


)
0=x cannot be the case because there is no such rectangle:


since we limited ourselves to the first quadrant.
To prove that this is indeed the maximum: We can say that for x=0 and x=3 our area is zero. If our x value is any number between 0 and 3 then our area is positive, our Area function is continous and differentiable and therefore by Rolle's Theorem (
://en.wikipedia.org/wiki/Rolle%27s_theorem) (Or common sense if you like) THere must be at least one maximum turning point on the interval (0,3) and we found that this can only occur at

, hence by deduction it is the maximum.
Note: that last paragraph is probably not completely required for specialist maths but I guess you will slowly get used to providing more rigorous proofs as you continue mathematics.
Oh and yeah, just sub in

into our Area function and double the value. Make sure to answer the question, that advice is applicable to any course
