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 2^nd 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 2^nd 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 .