Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2010 | June | Q#6

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Question

The diagram shows a block of wood in the shape of a prism with triangular cross-section. The end  faces are right-angled triangles with sides of lengths 3x cm, 4x cm and 5x cm, and the length of the  prism is y cm, as shown in the diagram.

The total surface area of the five faces is 144 cm2.

a.    

                           i.       Show that  .

                         ii.       Hence show that the volume of the block, V cm3, is given by

b.    

                           i.       Find .

                         ii.       Show that V has a stationary value when .

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

Solution

a.
 


i.
 

We are given that total surface area of the five faces is 144 cm2.

It is evident from the diagram that we have three rectangular and two triangular surfaces.

Therefore;

Expression for the area of the triangle is;

Expression for the area of the rectangle is;

Hence;

We can substitute given value of total surface area;


ii.
 

It is evident from the diagram that;

Expression for the volume of the rectangular block is;

We have found from (a:i) that;

We can rearrange the equation to find an expression for y;

Substituting tis value of y in expression of the volume found above;

b.
 


i.
 

We have found in (a:ii) that;

We are required to find;

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.

A stationary point  on the curve  is the point where gradient of the curve is equal to zero;

We have found in (b:ii) that;

Therefore, if V has a stationary value at point , then derivative must be equal to zero at this  point.

Hence, substituting  in expression of ;

Hence, V has a stationary value at .

c.
 

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 have found in (b:ii) that;

Therefore, we need second derivative of V.

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;

Therefore;

Rule for differentiation is of  is:

Rule for differentiation is of  is:

Rule for differentiation is of  is:

We are required to find the nature of stationary value at point .

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.

Therefore;

Since , V has a maximum value at point .

 

 

 

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