Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | May-Jun | (P1-9709/11) | Q#5

Question

A farmer divides a rectangular piece of land into 8 equal-sized rectangular sheep pens as shown in the diagram. Each sheep pen measures x m by y m and is fully enclosed by metal  fencing. The farmer uses 480m of fencing.

i.
Show that the total area of land used for the sheep pens, A m2, is given by

ii.       Given that x and y can vary, find the dimensions of each sheep pen for which the value of A  is a maximum. (There is no need to verify that the value of A is a maximum.)

Solution

i.

Expression for the area of the rectangle is;

It is evident from the diagram that;

There are total 08 number of pens, hence, total area;

Let us calculate perimeter of all pens.

Be careful!

We cannot calculate total perimeter by multiplying perimeter of each pen with 8 just like area  because there are common sides between adjacent pens which would be counted more than  once.

Consider the diagram below to find the perimeter of the all pens. Green arrows represent y whereas red arrows represent x.

Therefore, entire perimeter of all pens can be written as;

We are given that 480m of fencing was used which is equivalent to perimeter of all the pens.

Hence;

We have found area of the land as;

Since we are required to express area in terms of x, we can find expression of y from equation perimeter and substitute in equation of area.

Substituting in equation of area;

ii.

In (i) we have got expression for the area as;

If x and y are variables then maximum area A must be a stationary point of the equation for the  area.

A stationary point  on the curve  is the point where gradient of the curve is equal to zero;

Therefore, first we need gradient of the equation of area.

Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve  with respect to  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

Therefore, at point where area is maximum;

Hence, area is stationary (maximum) when .

At this point, we are required to find dimensions of each pen.

We have found that at maximum area A . We need value of y at this point.

We utilize the equation of y as found in(i) to find the value of y when  .

Substituting  in equation of y;

Hence, when Area A is maximum the dimensions of each pen area  and .