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Date May 2008 Marks available 5 Reference code 08M.2.sl.TZ1.5
Level SL only Paper 2 Time zone TZ1
Command term Calculate, Find, and Hence Question number 5 Adapted from N/A

Question

A closed rectangular box has a height y cm and width x cm. Its length is twice its width. It has a fixed outer surface area of 300 cm2 .

Factorise 3x2+13x10.

[2]
i.a.

Solve the equation 3x2+13x10=0.

[2]
i.b.

Consider a function f(x)=3x2+13x10 .

Find the equation of the axis of symmetry on the graph of this function.

[2]
i.c.

Consider a function f(x)=3x2+13x10 .

Calculate the minimum value of this function.

[2]
i.d.

Show that 4x2+6xy=300.

[2]
ii.a.

Find an expression for y in terms of x.

[2]
ii.b.

Hence show that the volume V of the box is given by V=100x43x3.

[2]
ii.c.

Find dVdx.

[2]
ii.d.

(i)     Hence find the value of x and of y required to make the volume of the box a maximum.

(ii)    Calculate the maximum volume.

[5]
ii.e.

Markscheme

(3x2)(x+5)     (A1)(A1)

[2 marks]

i.a.

(3x2)(x+5)=0

x=23 or x=5     (A1)(ft)(A1)(ft)(G2)

[2 marks]

i.b.

x=136 (2.17)     (A1)(A1)(ft)(G2)


Note: (A1) is for x=, (A1) for value. (ft) if value is half way between roots in (b).

[2 marks]

i.c.

Minimum y=3(136)2+13(136)10     (M1)

Note: (M1) for substituting their value of x from (c) into f(x) .


=24.1     (A1)(ft)(G2)

[2 marks]

i.d.

Area=2(2x)x+2xy+2(2x)y     (M1)(A1)


Note: (M1) for using the correct surface area formula (which can be implied if numbers in the correct place). (A1) for using correct numbers.

300=4x2+6xy     (AG)


Note:
Final line must be seen or previous
(A1) mark is lost.

[2 marks]

ii.a.

6xy=3004x2     (M1)

y=3004x26x or 1502x23x     (A1)

[2 marks]

ii.b.

Volume=x(2x)y     (M1)

V=2x2(3004x26x)     (A1)(ft)

=100x43x3     (AG)


Note: Final line must be seen or previous (A1) mark is lost.

[2 marks]

ii.c.

dVdx=10012x23  or  1004x2     (A1)(A1)

 Note: (A1) for each term.

[2 marks]

ii.d.

Unit penalty (UP) is applicable where indicated in the left hand column

(i)     For maximum dVdx=0  or  1004x2=0     (M1)

x=5     (A1)(ft)

y=3004(5)26(5)  or  (1502(5)23(5))     (M1)

=203     (A1)(ft)

(UP)     (ii)    33313 cm3 (333 cm3)


Note: (ft)
from their (e)(i) if working for volume is seen.

[5 marks]

ii.e.

Examiners report

Most candidates made a good attempt to factorise the expression.

i.a.

Many gained both marks here from a correct answer or ft from the previous part.

i.b.

Many used the formula correctly. Some forgot to put x= .

i.c.

Most candidates found this value from their GDC.

i.d.

A good attempt was made to show the correct surface area.

ii.a.

Many could rearrange the equation correctly.

ii.b.

Although this was not a difficult question it probably looked complicated for the candidates and it was often left out.

ii.c.

Those who reached this length could usually manage the differentiation.

ii.d.

(i)     Many found the correct value of x but not of y.

(ii)    This was well done and again the units were included in most scripts.

ii.e.

Syllabus sections

Topic 7 - Introduction to differential calculus » 7.6 » Optimization problems.
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