Mathematics · May / June 2025 · Paper 21 · Question 9
YMA01
As printed in the paper
The question as text
9. B C F D A E Figure 2 Figure 2 shows a sketch of an open container. The sides CDEF and ABFE are rectangles. The ends ADE and BCF are congruent (identical) right-angled triangles. The container is made from metal of negligible thickness. Given that • AE = BF = 3x metres • DE = CF = 2x metres • AB = DC = EF = L metres 3 and the capacity of the container is 12 m
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
9(a)[4]
Lx
2
x
x
20
2
2
=
6
+
5
=
6
+
S
x
xL
x
x
d
20
S
=
12
x
−
9(b)[5]
d
x
2
x
d
S
5
3
x =
0
=
=
1.18563...
x
x
d
3
20
2
=
6 ''1.18563''
×
+
=
25.3
(m2)
S
''1.18563''
x
=
1.18563
2
d
40
S
9(c)[2]
=
12
+
2
3
d
x
x
2
d
S
Justifies
(at
x =
1.18563...
), hence minimum *
>
0
2
d
x
(a)
M1: Attempts to use a volume formula to make L or Lx the subject. May be seen substituted into a
surface area expression which is not in the required form of the final answer.
β
β
12
=
α
2
=
Allow
x L
L
or Lx
=
2
x
x
4
4
=
A1:
L
OR
=
o.e. Maybe seen substituted into the equation for the surface area.
Lx
2
x
x
"4"
"4"
dM1: Substitutes
L
=
OR
into
2
which may be unsimplified to produce a
Lx
=
S
=
Rx
+
TxL
2
x
x
formula of the correct form. It is dependent on the previous method mark.
20
A1:
2
S
=
6
x
+
x
20
2
Must see “ S =” or e.g. “(surface) area = ” connected at some point with
6x
+
x
They may work algebraically using P and Q. They can still score M1A1ft in (b) but they
would have to substitute in their values to score any further marks in (b) or any marks in (c)
Condone other letters used instead of S and x for d
S
(b)
x including y for S.
d
Accept use of e.g. S′
d
S
b
M1: Correct form for the derivative
=
+
ax
d
x
2
x
Q
Q
2
A1ft: Follow through on the derivative of their
S
=
Px
+
2
Px
−
x
2
x
dM1: Solves d
S
x =
0
=
...
Do not be concerned by the mechanics of the rearrangement as long as a
x
d
value for x is achieved. =0 may be implied by further work or if just a correct value for x then you
may need to check this. It is dependent on the previous method mark.
ddM1: Substitutes their value of x in S to find S. Condone slips. It is dependent on the previous two method
marks.
If only a value is stated you may need to check this.
A1:
awrt 25.3 (m2) following correct differentiation. Units not required.
2
d
or d
S
(c)
Note the final A1* can only be scored using a correct expression for
d
2
d
x
2
d
S
on the approach taken). Condone other letters used instead of S and x for
2
d
x
S. Accept use of e.g. S′′
If second differentiation is seen in (b) then it must be seen or used in (c) to score.
2
d
S
N
M1: Differentiates and finds
of the form
M
+
(do not be concerned about signs for this mark).
2
x
3
d
x
Alternatively, substitutes in values of x above and below their 1.1856.. and proceeds to find values
for d
S
x at these points.
d
2
d
S
A1*: Either justifies
0
, which must be a correct expression AND gives a minimal conclusion
>
2
d
x
Requires one of:
2
d
S
a correct value for
(awrt 36 if using 1.18(56..) but allow awrt 35.1 if using 1.2) from a
2
d
x
correct expression, a correct comparison with 0 and a conclusion e.g. min
deduces that as 12 and 40 are both positive and the correct value of x (awrt 1.2)
2
d
S
>
0
with a conclusion e.g. proven
2
d
x
Or justifies there is a change of sign for d
S
x AND gives a minimal conclusion
d
Requires both calculations to be correct using a correct expression for d
S
x (you will need to
d
check this)
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