# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2010 | Oct-Nov | (P1-9709/12) | Q#10

Question

The diagram shows an open rectangular tank of height  meters covered with a lid. The base of the tank has sides of length  meters and  meters and the lid is a rectangle with sides of length  meters  meters. When full the tank holds 4m3 of water. The material from which the tank is made is of negligible thickness. The external surface area of the tank together with the area of the top of the lid is  m2.

i.
Express  in terms of  and hence show that  .

ii.       Given that  can vary, find the value of  for which  is a minimum, showing clearly that  is a minimum and not a maximum.

Solution

i.

We are given that when full the tank holds 4m3 of water. Hence, volume of the tank is;

Expression for the volume of the rectangular block is;

For the given case;

Hence;

Expression for the surface area of the rectangle is;

For the given case, top of the tank is not included and a rectangle of different dimensions is serving as the top.

First we find surface area of open tank.

Now we find surface area of rectangular lid.

Expression for the area of the rectangle is;

Hence, surface area of tank with lid;

We have found that;

Therefore;

ii.

A stationary value is the maximum or minimum value of a function. It can be obtained by equating the derivative of the function with zero. This gives the x-coordinates of the stationary point. At this point function has a stationary value.

For the given case;

Therefore, we need derivative of this function.

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

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Now we equate the derivative of the given function with ZERO.

At this point , area  has a stationary value.

We can determine nature of stationary point (value), whether minimum or maximum, by finding 2nd derivative of the curve.

Second derivative is the derivative of the derivative. If we have derivative of the curve   as  , then expression for the second derivative of the curve  is;

For the given case;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Rule for differentiation of  is:

We find the value of second derivative at the stationary point i.e. ,

If  or  then stationary point (or its value) is minimum.

If  or  then stationary point (or its value) is maximum.

Hence, the stationary value of area is a maximum.