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Date November 2015 Marks available 5 Reference code 15N.2.hl.TZ0.13
Level HL only Paper 2 Time zone TZ0
Command term Show that Question number 13 Adapted from N/A

Question

The following diagram shows a vertical cross section of a building. The cross section of the roof of the building can be modelled by the curve f(x)=30ex2400, where 20x20.

Ground level is represented by the x-axis.

Find f.

[4]
a.

Show that the gradient of the roof function is greatest when x =  - \sqrt {200} .

[3]
b.

The cross section of the living space under the roof can be modelled by a rectangle CDEF with points {\text{C}}( - a,{\text{ }}0) and {\text{D}}(a,{\text{ }}0), where 0 < a \le 20.

Show that the maximum area A of the rectangle CDEF is 600\sqrt 2 {{\text{e}}^{ - \frac{1}{2}}}.

[5]
c.

A function I is known as the Insulation Factor of CDEF. The function is defined as I(a) = \frac{{P(a)}}{{A(a)}} where {\text{P}} = {\text{Perimeter}} and {\text{A}} = {\text{Area of the rectangle}}.

(i)     Find an expression for P in terms of a.

(ii)     Find the value of a which minimizes I.

(iii)     Using the value of a found in part (ii) calculate the percentage of the cross sectional area under the whole roof that is not included in the cross section of the living space.

[9]
d.

Markscheme

f'(x) = 30{{\text{e}}^{ - \frac{{{x^2}}}{{400}}}} \bullet  - \frac{{2x}}{{400}}\;\;\;\left( { =  - \frac{{3x}}{{20}}{{\text{e}}^{ - \frac{{{x^2}}}{{400}}}}} \right)     M1A1

 

Note:     Award M1 for attempting to use the chain rule.

 

f''(x) =  - \frac{3}{{20}}{{\text{e}}^{ - \frac{{{x^2}}}{{400}}}} + \frac{{3{x^2}}}{{4000}}{{\text{e}}^{ - \frac{{{x^2}}}{{400}}}}\;\;\;\left( { = \frac{3}{{20}}{{\text{e}}^{ - \frac{{{x^2}}}{{400}}}}\left( {\frac{{{x^2}}}{{200}} - 1} \right)} \right)     M1A1

 

Note:     Award M1 for attempting to use the product rule.

[4 marks]

a.

the roof function has maximum gradient when f''(x) = 0     (M1)

 

Note:     Award (M1) for attempting to find f''\left( { - \sqrt {200} } \right).

 

EITHER

= 0     A1

OR

f''(x) = 0 \Rightarrow x =  \pm \sqrt {200}     A1

THEN

valid argument for maximum such as reference to an appropriate graph or change in the sign of f''(x) eg f''( - 15) = 0.010 \ldots ( > 0) and f''( - 14) =  - 0.001 \ldots ( < 0)     R1

\Rightarrow x =  - \sqrt {200}     AG

[3 marks]

b.

A = 2a \bullet 30{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}}\;\;\;\left( { = 60a{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}} =  - 400g'(a)} \right)     (M1)(A1)

EITHER

\frac{{{\text{d}}A}}{{{\text{d}}a}} = 60a{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}} \bullet  - \frac{a}{{200}} + 60{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}} = 0 \Rightarrow a = \sqrt {200} {\text{ }}\left( { - 400f''(a) = 0 \Rightarrow a = \sqrt {200} } \right)     M1A1

OR

by symmetry eg a =  - \sqrt {200} found in (b) or {A_{{\text{max}}}} coincides with f''(a) = 0     R1

\Rightarrow a = \sqrt {200}     A1

 

Note:     Award A0(M1)(A1)M0M1 for candidates who start with a = \sqrt {200} and do not provide any justification for the maximum area. Condone use of x.

 

THEN

{A_{{\text{max}}}} = 60 \bullet \sqrt {200} {{\text{e}}^{ - \frac{{200}}{{400}}}}     M1

= 600\sqrt 2 {{\text{e}}^{ - \frac{1}{2}}}     AG

[5 marks]

c.

(i)     perimeter = 4a + 60{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}}     A1A1

 

Note:     Condone use of x.

 

(ii)     I(a) = \frac{{4a + 60{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}}}}{{60a{{\text{e}}^{ - \frac{{{a^2}}}{{400}}}}}}     (A1)

graphing I(a) or other valid method to find the minimum     (M1)

a = 12.6     A1

(iii)     area under roof = \int_{ - 20}^{20} {30{{\text{e}}^{ - \frac{{{x^2}}}{{400}}}}} {\text{d}}x     M1

= 896.18 \ldots     (A1)

area of living space = 60 \cdot (12.6...) \cdot e - {\frac{{(12.6...)}}{{400}}^2} = 508.56...

percentage of empty space = 43.3\%     A1

[9 marks]

Total [21 marks]

d.

Examiners report

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d.

Syllabus sections

Topic 6 - Core: Calculus » 6.3 » Optimization problems.

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