Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | Feb-Mar | (P1-9709/12) | Q#6

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Question

A vacuum flask (for keeping drinks hot) is modelled as a closed cylinder in which the internal radius  is r cm and the internal height is h cm. The volume of the flask is 1000 cm3. A flask is most efficient when the total internal surface area, A cm2, is a minimum.

     i.       Show that  

   ii.       Given that r can vary, find the value of r, correct to 1 decimal place, for which A has a stationary  value and verify that the flask is most efficient when r takes this value.

Solution

a.
 

We are given that vacuum flask is cylinder and;

Expression for surface area of a cylinder is with radius  and height  is;

Therefore surface area of vacuum flask is;

We are required to express surface area of vacuum flask independent of .

Expression for volume of a cylinder is with radius  and height  is;

For the given case;

Substituting this expression for  in above expression of surface area of vacuum flask;


i.
 

We are given that A (surface area of vacuum falsk) has a stationary.

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.

From (ii) we have;

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

Therefore, we need;

Rule for differentiation of  is:

Rule for differentiation of  is:

Coordinates of stationary point on the curve  can be found from the derivative of equation of the  curve by equating it with ZERO. This results in value of x-coordinate of the stationary point  on the curve.

Therefore, value of  for which  has a stationary value can be found as follows;

Therefore, stationary value of  occurs when

We are given that flask is most efficient when the total internal surface area, A cm2, is a minimum.  We have found that flask has a stationary value of  when .

If this value of  is minimum then flask would be most efficient.

Once we have the coordinates of the stationary point  of a curve, we can determine its  nature, 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;

We substitute  of the stationary point in the expression of 2nd derivative of the curve and  evaluate it;

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

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

We have;

Therefore;

Rule for differentiation of  is:

Rule for differentiation of  is:

Since  has stationary value at ;

Since,  at , the  has a minimum value at this point and, therefore, flask will be most  efficient when .

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