Standard +0.8 This is a constrained optimization problem requiring calculus (differentiation, finding stationary points, and verification of minimum), combined with geometric reasoning about a cylinder. Students must set up the constraint V=ΟrΒ²h=8000, express total weld length L=2Οr+h as a function of one variable, differentiate, solve the resulting equation, and verify it's a minimum. The multi-step nature, need for algebraic manipulation, second derivative test, and justification of the answer pushes this above average difficulty, though it follows a standard optimization framework taught at AS level.
Rakti makes open-topped cylindrical planters out of thin sheets of galvanised steel.
She bends a rectangle of steel to make an open cylinder and welds the joint. She then welds this cylinder to the circumference of a circular base.
\includegraphics{figure_11}
The planter must have a capacity of \(8000\text{cm}^3\)
Welding is time consuming, so Rakti wants the total length of weld to be a minimum.
Calculate the radius, \(r\), and height, \(h\), of a planter which requires the minimum total length of weld.
Fully justify your answers, giving them to an appropriate degree of accuracy.
[9 marks]
Question 11:
11 | Obtains correct weld length in
terms of h and r | AO1.1b | B1 | Length of weld = w = h + 2Οr
Volume = 8000 = Οr2h
8000
So h =
ππ2
8000
w = + 2Οr
ππ2
For minimum length of weld
ππ€
= 0
ππ
β16000
+ 2Ο = 0
ππ3
leading to Ο2r3 = 8000
r = 9.32 cm
h = 29.3 cm
π2π€ 48000
=
ππ2 ππ4
which is positive, so this is a
minimum for w
Obtains formula for h in terms of r
or vice versa using volume = 8000 | AO3.1b | M1
Substitutes to get weld length in
terms of one variable, obtaining
correct formula for w | AO1.1b | A1
States that for a stationary point
the first derivative is zero. (OE) | AO2.4 | E1
Differentiates correctly (FT
provided formula includes negative
powers) (accept numerical value of
16000
β used)
π | AO1.1b | B1F
Solves to find a value of r and a
value of h (do not award if the final
value of r or h is negative) | AO1.1a | M1
Obtains r = 9 or 9.3 (AWRT) and
h = 29, 29.3 (AWRT) or 30 | AO1.1b | A1
Differentiates βtheirβ first differential
and substitutes in βtheirβ value of r
or h | AO1.1a | M1
Sets out a well-constructed
mathematical argument, using
precise statements throughout to
find the values of r (9 or 9.3) and h
(29, 29.3 or 30) and justifies the
minimum value. Can be awarded if
E1 not obtained. | AO2.1 | R1
Total | 9
Q | Marking Instructions (II) | AO | Marks | Typical Solution (using h)
11 | Length of weld = w = h + 2Οr
Volume = 8000 = Οr2h
8000
r = β
πβ
8000
w = h + 2Οβ
πβ
1
= β+ β32000πβ β 2
For minimum length of weld
ππ€
= 0
πβ
1 β 3
1β β32000πβ 2 = 0
2
leading to h3 = 8000Ο
h = 29.3 cm
r = 9.32 cm
π2π€ 3 5
β
= β32000πβ 2
πβ2 4
which is positive, so this is a
minimum for w
Total | 9
Q | Marking Instructions | AO | Marks | Typical Solution
Rakti makes open-topped cylindrical planters out of thin sheets of galvanised steel.
She bends a rectangle of steel to make an open cylinder and welds the joint. She then welds this cylinder to the circumference of a circular base.
\includegraphics{figure_11}
The planter must have a capacity of $8000\text{cm}^3$
Welding is time consuming, so Rakti wants the total length of weld to be a minimum.
Calculate the radius, $r$, and height, $h$, of a planter which requires the minimum total length of weld.
Fully justify your answers, giving them to an appropriate degree of accuracy.
[9 marks]
\hfill \mbox{\textit{AQA AS Paper 2 2018 Q11 [9]}}