Graphs 4
4 marks · HardA rectangular field is to be enclosed on three sides with fencing. The fourth side is a hedge and requires no fence. The length of the field is metres and the width is metres. If 160 m of fencing is available, express the total area of the field in terms of , and find the value of that gives the maximum area.
Show that .
Calculate the value of that produces the largest field area, and find this maximum area.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) A = 160x - 2x^2
(b) x = 40 m; maximum area = 3200 m^2
Must show: Correct expression for area, Maximum via axis of symmetry/formula/substitution
Common mistakes: Dividing by 2 or 3 incorrectly; Using all four sides for the perimeter
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Correctly shows/factors A = 160x - 2x^2.
M1 (b) · 1 mark · Correct method: uses axis of symmetry or vertex formula to maximize area.
A1 (b) · 1 mark · Correct value of x = 40 m. (follow-through)
A2 (b) · 1 mark · Correct maximum area: 3200 m^2. (follow-through)
Worked solution
Let y = length; x = width. 3 sides are fenced: y + 2x = 160 → y = 160 - 2x. Area A = x \times y = x(160 - 2x) = 160x - 2x^2.
(a) Shown: A = 160x - 2x^2.
(b) Max area when x = -b/(2a): a = -2, b = 160 → x = -160/(2\times -2) = 40.
At x = 40, y = 160 - 2\times 40 = 80 m; area = 40 \times 80 = 3200 m^2.