Find $f''(x)$.

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

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 $600\sqrt 2 {{\text{e}}^{ - \frac{1}{2}}}$.

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.

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

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]

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

(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]