Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2017 | June | Q#7

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Question

The diagram shows the right-angled corner AFE of a building and four sections of fencing running  parallel to the walls of the building.

Each of the sections of fencing AB and DE has length x metres and each of the sections of wall AF  and FE has length y metres. The total length of the four sections of fencing AB, BC, CD and DE is  15 metres. The shaded region bounded by the fencing and the walls of the building has area S m2

A circle with centre C has .

a.    

                    i.       Express y in terms of x.

                  ii.       Show that

b.    

                    i.       Express  in the form , where p and q are rational numbers.

                  ii.       Hence find the maximum value of S.

Solution

a.
 

                            i.
 

We are given that the total length of the four sections of fencing AB, BC, CD and DE is 15 metres. 

Therefore;

It is evident from the diagram that;

Hence;

                          ii.
 

We are required to obtain expression for the area of shaded region S.

Consider the diagram below.

It is evident from the diagram that;

Expression for the area of the rectangle is;

Therefore;

As we have seen above;

From (a:i) we have;

Therefore;

b.
 

                        i.
 

We have the expression;

We use method of “completing square” to obtain the desired form.

Next we complete the square for the terms which involve .

We have the algebraic formula;

For the given case we can compare the given terms with the formula as below;

Therefore we can deduce that;

Hence we can write;

To complete the square we can add and subtract the deduced value of ;

                          ii.
 

We have found in (a:ii) that;

From (b:i) we know that;

Therefore;

Standard form of quadratic equation is;

The graph of quadratic equation is a parabola. If  (‘a’ is positive) then parabola opens upwards  and its vertex is the minimum point on the graph.
If
 (‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the  graph.

We recognize that given curve , is a parabola opening downwards.

Vertex form of a quadratic equation is;

As demonstrated in a(:ii);

Coordinates of the vertex are .Since this is a parabola opening upwards the vertex is the  minimum point on the graph.
Here y-coordinate of vertex represents least value of
 and x-coordinate of vertex represents  corresponding value of .

For the given case, vertex is . Therefore, maximum value of  is  and corresponding value
of
 is .

However, we are required to find the maximum value of S;

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