Show that the area, $A\;{\text{c}}{{\text{m}}^2}$, of the triangle is given by $A = \frac{1}{4}({x^3} - 8{x^2} + 5x + 50)$.

(i)     State $\frac{{{\text{d}}A}}{{{\text{d}}x}}$.

(ii)     Verify that $\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0$ when $x = \frac{1}{3}$.

(i)     Find $\frac{{{{\text{d}}^2}A}}{{{\text{d}}{x^2}}}$ and hence justify that $x = \frac{1}{3}$ gives the maximum area of triangle $PQR$.

(ii)     State the maximum area of triangle $PQR$.

(iii)     Find $QR$ when the area of triangle $PQR$ is a maximum.

use of $A = \frac{1}{2}qr\sin \theta$ to obtain $A = \frac{1}{2}(x + 2){(5 - x)^2}\sin 30^\circ$     M1

$= \frac{1}{4}(x + 2)(25 - 10x + {x^2})$     A1

$A = \frac{1}{4}({x^3} - 8{x^2} + 5x + 50)$     AG

[2 marks]

(i)     $\frac{{{\text{d}}A}}{{{\text{d}}x}} = \frac{1}{4}(3{x^2} - 16x + 5) = \frac{1}{4}(3x - 1)(x - 5)$     A1

(ii)     METHOD 1

EITHER

$\frac{{{\text{d}}A}}{{{\text{d}}x}} = \frac{1}{4}\left( {3{{\left( {\frac{1}{3}} \right)}^2} - 16\left( {\frac{1}{3}} \right) + 5} \right) = 0$     M1A1

OR

$\frac{{{\text{d}}A}}{{{\text{d}}x}} = \frac{1}{4}\left( {3\left( {\frac{1}{3}} \right) - 1} \right)\left( {\left( {\frac{1}{3}} \right) - 5} \right) = 0$     M1A1

THEN

so $\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0$ when $x = \frac{1}{3}$     AG

METHOD 2

solving $\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0$ for $x$     M1

$- 2 < x < 5 \Rightarrow x = \frac{1}{3}$     A1

so $\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0$ when $x = \frac{1}{3}$     AG

METHOD 3

a correct graph of $\frac{{{\text{d}}A}}{{{\text{d}}x}}$ versus $x$     M1

the graph clearly showing that $\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0$ when $x = \frac{1}{3}$     A1

so $\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0$ when $x = \frac{1}{3}$     AG

[3 marks]

(i)     $\frac{{{{\text{d}}^2}A}}{{{\text{d}}{x^2}}} = \frac{1}{2}(3x - 8)$     A1

for $x = \frac{1}{3},{\text{ }}\frac{{{{\text{d}}^2}A}}{{{\text{d}}{x^2}}} =  - 3.5{\text{ }}( < 0)$     R1

so $x = \frac{1}{3}$ gives the maximum area of triangle $PQR$     AG

(ii)     ${A_{\max }} = \frac{{343}}{{27}}{\text{ }}( = 12.7){\text{ (c}}{{\text{m}}^2}{\text{)}}$     A1

(iii)     ${\text{PQ}} = \frac{7}{3}{\text{ (cm)}}$ and ${\text{PR}} = {\left( {\frac{{14}}{3}} \right)^2}{\text{ (cm)}}$     (A1)

${\text{Q}}{{\text{R}}^2} = {\left( {\frac{7}{3}} \right)^2} + {\left( {\frac{{14}}{3}} \right)^4} - 2\left( {\frac{7}{3}} \right){\left( {\frac{{14}}{3}} \right)^2}\cos 30^\circ$     (M1)(A1)

$= 391.702 \ldots$

${\text{QR = 19.8 (cm)}}$     A1

[7 marks]

Total [12 marks]

This question was generally well done. Parts (a) and (b) were straightforward and well answered.

This question was generally well done. Parts (a) and (b) were straightforward and well answered.

This question was generally well done. Parts (c) (i) and (ii) were also well answered with most candidates correctly applying the second derivative test and displaying sound reasoning skills.

Part (c) (iii) required the use of the cosine rule and was reasonably well done. The most common error committed by candidates in attempting to find the value of $QR$ was to use ${\text{PR}} = \frac{{14}}{3}{\text{ (cm)}}$ rather than ${\text{PR}} = {\left( {\frac{{14}}{3}} \right)^2}{\text{ (cm)}}$. The occasional candidate used $\cos 30^\circ  = \frac{1}{2}$.