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Question

Q1. A farmer uses 1000 metres of fencing to construct a rectangular-shaped field. His land is located alongside a river, so he only needs to fence three sides of the field. There is also a rectangular-shaped dam of area 5000 m² within the field.

Let x be the width of the two shorter sides of the field.

a. If y is the length of the field, express y in terms of x. Let A m² be the area of the farmable land inside the fence.

b. Show that A= -2x^{2 +} 1000x - 5000

c. Sketch the graph of A= -2x^{2 +} 1000x - 5000

d. Use your calculator to find the coordinates of the turning point of the graph.

e. Find the dimensions of the field which gives the maximum farmable area, and state the maximum area?

f. Either by completing the square or otherwise, express the rule for A in the form A= a(x-h)²+k

Q2. The number of tables N sold by an outdoor furniture store during a particular year can be

approximated by the equation: N= m² -14m+71 where m is the number of months into the year, m ∈ [1,12].

a. Sketch the graph of the sales for the store over 12 months

b. How many tables did the store sell in March?

c. In which months did the store sell 47 tables?

d. What is the minimum number of tables sold in a month and in which month did this happen?

The store needs to sell at least 24 tables in a month in order to make a profit.

e. For how many months of the year did the store make a profit?

[Don't need to sketch graphs]

Answer 1

1000=L +2w

L = 1000-2w

or if x=width

then

A= -2x^2 +1000x

and subtract the area of the non farmable land of 5000 m^2

A=-2x^2 +1000x -5000=xy

y = -2x+1000-5000/x

Area = A =Lw = (1000-2w)w= 1000w -2w^2

A'(w) = 1000-4w = 0

4w = 1000

w =x= 1000/4 =250 meters wide

L =y= 1000-2x= 1000-2(250) = 1000-500 = 500 meters Long

are the dimensions to maximize area given 1000 meters of fencings

max Area of farmable land = Lw -5000=yx-5000= (500)(250)-5000 = 125,000 m^2-5000 m^2

= 120,000 m^2

the graph of A is a downward opening parabola with vertex = maximum = (250, 120,000)

the turning point is the vertex where A'=0

A=-2x^2 +1000x -5000

=-2(x^2 -500x) -5000

= -2(x^2 -500x +250^2) -5000+2(250^2)

=-2(x-250)^2 -5000+2(62,500)

=-2(x-250)^2 -5,000+125,000

=-2(x-250)^2 +120,000

=a(x-h)^2 +k

where (h,k) = the vertex = (250,120000)

and a=-2