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Date November Example questions Marks available 2 Reference code EXN.1.SL.TZ0.7
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Find Question number 7 Adapted from N/A

Question

The following diagram shows the graph of y=4-x2, 0x2 and rectangle ORST. The rectangle has a vertex at the origin O, a vertex on the y-axis at the point R0,y, a vertex on the x-axis at the point Tx,0 and a vertex at point Sx,y on the graph.

Let P represent the perimeter of rectangle ORST.

Let A represent the area of rectangle ORST.

Show that P=-2x2+2x+8.

[2]
a.

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

[6]
b.

Find an expression for A in terms of x.

[2]
c.

Find the dimensions of rectangle ORST that has maximum area.

[5]
d.

Determine the maximum area of rectangle ORST.

[1]
e.

Markscheme

* 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=2x+2y        (A1)

=2x+24-x2        A1

so  P=-2x2+2x+8        AG

 

[2 marks]

a.

METHOD 1

EITHER

uses the axis of symmetry of a quadratic        (M1)

x=-22-2

 

OR

forms dPdx=0        (M1)

-4x+2=0

 

THEN

x=12        A1

substitutes their value of x into y=4-x2        (M1)

y=4-122

y=154        A1

so the dimensions of rectangle ORST of maximum perimeter are 12 by 154

 

EITHER

substitutes their value of x into P=-2x2+2x+8        (M1)

P=-2122+212+8

 

OR

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

P=212+2154

P=172        A1

so the maximum perimeter is 172

 

METHOD 2

attempts to complete the square         M1

P=-2x-122+172        A1

x=12        A1

substitutes their value of x into y=4-x2        (M1)

y=4-122

y=154        A1

so the dimensions of rectangle ORST of maximum perimeter are 12 by 154

P=172        A1

so the maximum perimeter is 172        

 

[6 marks]

b.

substitutes y=4-x2 into A=xy        (M1)

A=x4-x2  =4x-x3        A1

 

[2 marks]

c.

dAdx=4-3x2        A1

attempts to solve their dAdx=0 for x        (M1)

4-3x2=0

x=23=233 x>0        A1

substitutes their (positive) value of x into y=4-x2        (M1)

y=4-232

y=83        A1

 

[5 marks]

d.

A=1633=1639        A1

 

[1 mark]

e.

Examiners report

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a.
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b.
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c.
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d.
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e.

Syllabus sections

Topic 5 —Calculus » SL 5.8—Testing for max and min, optimisation. Points of inflexion
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Topic 5 —Calculus

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