Question 9:
--- 9(a) ---
9(a) | Identifies and clearly defines
variable for radius of cylinder.
Can be shown on diagram or
can be implied by use in
V =πr2h | 2.5 | B1 | Radius of cylinder = r
h2 +r2 = R2
V =πr2h
( )
V =π R2 −h2 h
=πR2h−πh3
Uses Pythagoras to connect h, r
and R | 3.1a | M1
Eliminates the radius variable to
form an expression for the
volume of the cylinder in terms
of h, completing argument to
show given result. Condone
undefined r | 2.1 | R1
Subtotal | 3
--- 9(b) ---
9(b) | Differentiates the expression for
volume w.r.t. h with at least one
term correct. | 3.1a | M1 | dV
=πR2 −3πh2
dh
dV
For maximum volume =0
dh
⇒ R2 −3h2 =0
R2 R
h2 = ⇒h=
3 3
Hence volume
R  R 
V =πR2 −π  
3  3
2 3πR3
=
9
d2V
=−6πh
dh2
R
h=
When
3
d2V
<0Therefore
maximum
dh2
dV
Obtains correct
dh | 1.1b | A1
Explains that their derivative
w.r.t h equals zero for a
maximum or stationary point | 2.4 | E1
Equates volume derivative w.r.t.
h to zero and correctly obtains a
value for h in terms of R | 1.1a | M1
Substitutes their h, in terms of R,
from derivative w.r.t. h into
volume formula. | 1.1a | M1
Obtains the correct max volume
kR3− pR3
in the form or better | 3.2a | A1
Justifies correct volume in the
kR3− pR3
form or better form is
the maximum
eg:
• V = 0 when h=0 or R and
V>0 in between.
• Sketches shape of graph
passing through the
origin with (min on
negative side) and max
on positive side
d2V
• Obtains =−6πh<0
dh2
NB R1 can be awarded even if
E1 is not awarded. | 2.1 | R1
Subtotal | 7
Question Total | 10
Q | Marking Instructions | AO | Marks | Typical Solution