A cuboid has a rectangular cross-section where the length of the rectangle is equal to twice its width, $x$ cm, as shown in Figure 2. The volume of the cuboid is 81 cubic centimetres. a) Show that the total length, $L$ cm, of the twelve edges of the cuboid is given by $$L = 12x + \frac{162}{x^2}$$. b) Use calculus to find the minimum value of $L$. c) Justify, by further differentiation, that the value of $L$ that you have found is a minimum.

A-Level Edexcel Maths Pure Integration: A cuboid has a rectangular cross

A cuboid has a rectangular cross-section where the length of the rectangle is equal to twice its width, $x$ cm, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 2

Question 9

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A cuboid has a rectangular cross-section where the length of the rectangle is equal to twice its width, $x$ cm, as shown in Figure 2. The volume of the cuboid is 81 ... show full transcript

Worked Solution & Example Answer:A cuboid has a rectangular cross-section where the length of the rectangle is equal to twice its width, $x$ cm, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 2

Step 1

Show that the total length, L cm, of the twelve edges of the cuboid is given by

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Answer

To find the expression for the total length, $L$ , we start with the formula for volume of a cuboid:

$V = ext{length} \times ext{width} \times ext{height}$

Given that the length is $2x$ , width is $x$ , and the height can be derived from the volume:

$81 = (2x)(x)(h)$

From this, we can express $h$ :

$h = \frac{81}{2x^2}$

The total length of all edges, $L$ , can be computed as:

$L = 4 \times ( ext{length} + ext{width} + ext{height}) = 4(2x + x + h) = 12x + 4h$

Substituting the value of $h$ in:

$L = 12x + 4 \left( \frac{81}{2x^2} \right) = 12x + \frac{162}{x^2}$

Step 2

Use calculus to find the minimum value of L.

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Answer

To find the minimum value of $L$ , we take the derivative of $L$ with respect to $x$ :

$\frac{dL}{dx} = 12 - \frac{324}{x^3}$

Setting the derivative equal to zero for critical points:

$12 - \frac{324}{x^3} = 0$

This leads to:

\therefore x = 3$$ Now, substituting $x = 3$ back into $L$ for the minimum value: $$L = 12(3) + \frac{162}{3^2} = 36 + 18 = 54$$

Step 3

Justify, by further differentiation, that the value of L that you have found is a minimum.

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Answer

To confirm that $x = 3$ gives a minimum for $L$ , we will look at the second derivative:

$\frac{d^2L}{dx^2} = \frac{972}{x^4}$

Since $x > 0$ , we have:

$\frac{d^2L}{dx^2} > 0$

This indicates that $L$ is concave up at $x = 3$ , confirming that this critical point is indeed a minimum.

A cuboid has a rectangular cross-section where the length of the rectangle is equal to twice its width, $x$ cm, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 2

Worked Solution & Example Answer:A cuboid has a rectangular cross-section where the length of the rectangle is equal to twice its width, $x$ cm, as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 2

Show that the total length, L cm, of the twelve edges of the cuboid is given by

Use calculus to find the minimum value of L.

Justify, by further differentiation, that the value of L that you have found is a minimum.

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