# Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2007 | January | Q#5

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Question

The diagram shows an open-topped water tank with a horizontal rectangular base and four vertical  faces. The base has width metres and length metres, and the height of the tank is metres. The combined internal surface area of the base and four vertical faces is 54m2.

a.

i.       Show that .

ii.       Hence express in terms of .

iii.       Hence show that the volume of water, V m3 , that the tank can hold when full is given by b.

i.       Find ii.       Verify that V has a stationary value when .

iii.        Find and hence determine whether V has a maximum value or a minimum value when .

Solution

a.

i.

Expression for the surface area of the box is;  Since given water tank is open-topped box, we need to modify the expression for its surface area as  follows;    We are given that; Comparing the two equation of surface area of open-topped box we get;    ii.

From a(i) we have; We can rearrange this equation to express in terms of .    iii.

Expression for the volume of the rectangular block is; For the given water tank (volume does not change with open top);   From a(ii) we have; We can substitute this value of in the expression of volume of water tank.     b.

i. From (a:iii) we have; Therefore; Rule for differentiation is of is:  Rule for differentiation is of is:     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. From (b:i) we have; Therefore;        Since represents width of the water tank and width cannot be negative; Hence has a stationary value at .

iii. From (b:i) we have; Therefore; Rule for differentiation is of is:  Rule for differentiation is of is: Rule for differentiation is of is:     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.

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.

For the given case . Substituting this value in ;  Since , stationary point at is maximum.