Maths

The Question says to justify the value of v in part b is the maximum. I got 6332 as my answer in b. So do I just state,
V= 6332
6332 > 0 hence the value is a maximum of V.

You may need a bit more justification than that (all volumes will be >=0), though the question is a bit vague about what they mean.
How do you normally justify whether a stationary point is a min or max? Even if it wasnt totally required for the question, it could come you in your as exams.

by stating whether it is Greater or Less than 0 right? and if it is equal to 0 then it could be both or a infection. Atleast thats what I see in the textbook

Yes, so just differentiate again and if the second derivaitve is negative (you expect it to be) its a maximum. Should only be a couple of lines to do it.

Ah I didn't do the second derivative part.
Just did it and I got a negative value of -10√6,
So I stated when x=16.3, d2y/dx2= -10√6 which is <0 hence this is a maximum point.

Is this enough?

The max should occur about x=23.1 (40/sqrt(3)). The corresponding V is slightly too high.
You may need to check that part, but apart from that looks good.

oh hmm. For the first differentiation I did,
400 - 3/4x^2 = 0
400 = 3/4x^2
x^2 = 400/ 3/4
x = √400/ 3/4
x = 20√6/3 = 16.3

Then I put it in my second derivative which was,
d2y/dx2 = -3/2(16.3)
= -10√6

400 / (3/4) = ...
It may be esaier to simply multiply through the previus equation by 4/3?