EXN.1.SL.TZ0.7

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The following diagram shows the graph of $y = 4 - x^{2}$ , $0 \leq x \leq 2$ and rectangle $ORST$ . The rectangle has a vertex at the origin $O$ , a vertex on the $y$ -axis at the point $R (0, y)$ , a vertex on the $x$ -axis at the point $T (x, 0)$ and a vertex at point $S (x, y)$ on the graph.

Let $P$ represent the perimeter of rectangle $ORST$ .

Let $A$ represent the area of rectangle $ORST$ .

Show that $P = - 2 x^{2} + 2 x + 8$ .

[2]

Find the dimensions of rectangle $ORST$ that has maximum perimeter and determine the value of the maximum perimeter.

[6]

Find an expression for $A$ in terms of $x$ .

[2]

Find the dimensions of rectangle $ORST$ that has maximum area.

[5]

Determine the maximum area of rectangle $ORST$ .

[1]

Markscheme / solution

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

$P = 2 x + 2 y$ (A1)

$= 2 x + 2 (4 - x^{2})$ A1

so $P = - 2 x^{2} + 2 x + 8$ AG

[2 marks]

METHOD 1

EITHER

uses the axis of symmetry of a quadratic (M1)

$x = - \frac{2}{2 (- 2)}$

forms $\frac{d P}{d x} = 0$ (M1)

$- 4 x + 2 = 0$

THEN

$x = \frac{1}{2}$ A1

substitutes their value of $x$ into $y = 4 - x^{2}$ (M1)

$y = 4 - {(\frac{1}{2})}^{2}$

$y = \frac{15}{4}$ A1

so the dimensions of rectangle $ORST$ of maximum perimeter are $\frac{1}{2}$ by $\frac{15}{4}$

EITHER

substitutes their value of $x$ into $P = - 2 x^{2} + 2 x + 8$ (M1)

$P = - 2 {(\frac{1}{2})}^{2} + 2 (\frac{1}{2}) + 8$

substitutes their values of $x$ and $y$ into $P = 2 x + 2 y$ (M1)

$P = 2 (\frac{1}{2}) + 2 (\frac{15}{4})$

$P = \frac{17}{2}$ A1

so the maximum perimeter is $\frac{17}{2}$

METHOD 2

attempts to complete the square M1

$P = - 2 {(x - \frac{1}{2})}^{2} + \frac{17}{2}$ A1

$x = \frac{1}{2}$ A1

substitutes their value of $x$ into $y = 4 - x^{2}$ (M1)

$y = 4 - {(\frac{1}{2})}^{2}$

$y = \frac{15}{4}$ A1

so the dimensions of rectangle $ORST$ of maximum perimeter are $\frac{1}{2}$ by $\frac{15}{4}$

$P = \frac{17}{2}$ A1

so the maximum perimeter is $\frac{17}{2}$

[6 marks]

substitutes $y = 4 - x^{2}$ into $A = x y$ (M1)

$A = x (4 - x^{2}) (= 4 x - x^{3})$ A1

[2 marks]

$\frac{d A}{d x} = 4 - 3 x^{2}$ A1

attempts to solve their $\frac{d A}{d x} = 0$ for $x$ (M1)

$4 - 3 x^{2} = 0$

$\Rightarrow x = \frac{2}{\sqrt{3}} (= \frac{2 \sqrt{3}}{3}) (x > 0)$ A1

substitutes their (positive) value of $x$ into $y = 4 - x^{2}$ (M1)

$y = 4 - {(\frac{2}{\sqrt{3}})}^{2}$

$y = \frac{8}{3}$ A1

[5 marks]

$A = \frac{16}{3 \sqrt{3}} (= \frac{16 \sqrt{3}}{9})$ A1

[1 mark]

Examiners’ report

[N/A]