Question 14:
--- 14(a) ---
14(a) | Identifies and clearly defines
variables. | AO3.1b | B1 | Let
C = total cost
x = length of base edges
h = length of height
C 15x232xh
x2h60
60
h
x2
1920
C 15x2  (*)
x
Differentiating
1920
30x 0
x2
x3 = 64
x = 4
60
h   3.75
42
Height of tank is 3.75 m
d2C 3840
30
dx2 x3
d2C
x4  >0 therefore minimum
dx2
ALT from (*) onwards
1
C 900h132 60h2
1

0900h2 16 60h 2
Height of tank is 3.75 m
d2C 3

1800h3 8 60h 2
dh2
d2C
h3.75 25.60
dh2
 minimum
Models the cost with an expression
of the form ax2 bxh | AO3.3 | M1
Eliminates either variable, using
volume equation, to form a model for
the cost in one variable. | AO3.3 | M1
Obtains a correct equation to model
cost in one variable | AO3.1b | A1
Uses their model to find minimum.
(at least one term correctly
differentiated and equated to zero) | AO3.4 | M1
Obtains correct equation | AO1.1b | A1
Obtains correct value for h with
correct units in context
Award FT from correct substitution
into incorrect equation for h but only
if all three M1 marks have been
awarded, must have correct units. | AO3.2a | A1F
Performs a correct test of ‘their’
solution: uses the second derivative
of ‘their’ expression for C in terms of
x or h to justify that a minimum value
for h has been found OE
(Second derivative > 0 or test
gradient/values either side) | AO2.4 | R1
Q | Marking Instructions | AO | Marks | Typical Solution
--- 14(b)(i) ---
14(b)(i) | Explains that two of the sides of the
tank will need to have length
x0.05 in order to join them | AO3.5c | E1 | The sides will need to overlap to be
joined, so two of the side lengths will
need to be x0.05
(b)(ii) | Explains that the refinement is
relatively small and unlikely to have
a significant effect on the result. | AO3.5a | R1 | The minimum cost is likely to
increase slightly, but relative to the
size of the tank, an extra 5cm is
unlikely to make a significant
difference.
Total | 10
Q | Marking Instructions | AO | Marks | Typical Solution