Or, in the spirit of applied differentiation:
restart:
Since h = 6 - r by the constraint r + h = 6 we have
V:=Pi*r^2*(6-r);
as you have already given.
If the max occurs on the open interval (0,6), then dV/dr = 0. A quick look at the plot:
plot(V,r=-1..7);
Solve dV/dr = 0 for r
diff(V,r);
solve(diff(V,r)=0,r);
So the maximum V occurs when r = 4.
MaxVol:=subs(r=4,V);
Thomas

Some additional information at

wikipedia-quantile. Also, some reference to the theory regarding computation at

Quantile_function linked to that same page at the bottom.

Some additional information at

wikipedia-quantile. Also, some reference to the theory regarding computation at

Quantile_function linked to that same page at the bottom.

Just luck, I had come across it by chance a few months ago. Mma uses the same term.
In wonder how it works as well. However, that is a topic you are eminently more qualified to discuss than I am. So, if it takes place, I will just have to read along and learn.

Just luck, I had come across it by chance a few months ago. Mma uses the same term.
In wonder how it works as well. However, that is a topic you are eminently more qualified to discuss than I am. So, if it takes place, I will just have to read along and learn.

with(Statistics):
df := 29:
chiSquare := RandomVariable(ChiSquare(df)):
Quantile(chiSquare, .95);
42.55696751
CDF(chiSquare, 42.55696751);
0.950000000033921820

with(Statistics):
df := 29:
chiSquare := RandomVariable(ChiSquare(df)):
Quantile(chiSquare, .95);
42.55696751
CDF(chiSquare, 42.55696751);
0.950000000033921820

Two comments:
First, you are missing a multiplication operator.
3x^2 should be 3*x^2
Second, if you are going to use "implicit" plot you need a second variable that is an
implicit function of x.
g1:= implicitplot(y-3*x^2=0, x=0..10, y=0..300);
display(g1);
or more simply
implicitplot(y-3*x^2=0, x=0..10, y=0..300);
But this is unnecessary since you can just use "plot" without y.
plot(3*x^2, x=0..10, y=0..300);

Two comments:
First, you are missing a multiplication operator.
3x^2 should be 3*x^2
Second, if you are going to use "implicit" plot you need a second variable that is an
implicit function of x.
g1:= implicitplot(y-3*x^2=0, x=0..10, y=0..300);
display(g1);
or more simply
implicitplot(y-3*x^2=0, x=0..10, y=0..300);
But this is unnecessary since you can just use "plot" without y.
plot(3*x^2, x=0..10, y=0..300);

I think about or getassumptions makes it more clear. That is why I put them in above.
getassumptions(S);
{Pi::Pi, _B1::(OrProp(0, 1)), _Z1::integer}
about(_Z1);
Originally _Z1, renamed _Z1~:
is assumed to be: integer
about(_B1);
Originally _B1, renamed _B1~:
is assumed to be: OrProp(0,1)
There is nothing to indicate that _Z1 and _B1 cannot both be 0. Maple gave the same solution regardless of whether or not I included x<>0. I only put that in because it had not been discussed in the previous posts and had been incorrectly included in the solution set. I suppose that one can infer that if it is input to solve directly then _Z1 and _B1 both equal to 0 is eliminated by the statement of the problem, but this is probably a stretch.

I think about or getassumptions makes it more clear. That is why I put them in above.
getassumptions(S);
{Pi::Pi, _B1::(OrProp(0, 1)), _Z1::integer}
about(_Z1);
Originally _Z1, renamed _Z1~:
is assumed to be: integer
about(_B1);
Originally _B1, renamed _B1~:
is assumed to be: OrProp(0,1)
There is nothing to indicate that _Z1 and _B1 cannot both be 0. Maple gave the same solution regardless of whether or not I included x<>0. I only put that in because it had not been discussed in the previous posts and had been incorrectly included in the solution set. I suppose that one can infer that if it is input to solve directly then _Z1 and _B1 both equal to 0 is eliminated by the statement of the problem, but this is probably a stretch.

My apologies for giving the wrong information. I should have looked more closely at the original post. Thanks Jacques for the correction.
Thomas

My apologies for giving the wrong information. I should have looked more closely at the original post. Thanks Jacques for the correction.
Thomas

Congratulations on 700, and thanks for putting that much time and effort into helping others at this site. It is greatly appreciated.
Thomas

Daniel,
I'm glad the links were helpful. I can't comment beyond that because I am not knowledgeable enough in these areas. Fortunately, Jacques was able to respond before I read your post.
Thomas