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

Hits: 287

**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 cm^{2}.

**a. **

** i. **Show that .

** ii. **Hence show that the volume of the block, V cm^{3}, 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 cm^{2}.

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 2^{nd }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 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.

Therefore;

Since , V has a maximum value at point .

## Comments