# 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

**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 cm^{3}. A flask is most efficient when the total internal surface area, A cm^{2}, 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 cm^{2}, 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 .

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