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.

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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 - Edexcel - A-Level Maths Pure - Question 1 - 2011 - Paper 2

Question 1

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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 1 - 2011 - Paper 2

Step 1

Show that the total length, $L$ cm, of the twelve edges of the cuboid is given by $L = 12x + \frac{162}{x^2}$

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Answer

To find the expression for the total length of the cuboid, we start with the volume formula.

The volume $V$ is given by the formula: $V = \text{length} \times \text{width} \times \text{height}$ Here, the length of the cuboid is $2x$ , the width is $x$ , and the height can be denoted as $h$ . Hence, $V = (2x) \cdot x \cdot h = 2x^2h$ Since the volume is given as 81 cubic centimetres, we have: $2x^2h = 81$ Solving for $h$ gives: $h = \frac{81}{2x^2}$

The total length, $L$ , of the twelve edges can be calculated using: $L = 4(\text{length}) + 4(\text{width}) + 4(\text{height}) = 4(2x) + 4(x) + 4(h)$ Simplifying this results in: $L = 8x + 4x + 4h = 12x + 4\frac{81}{2x^2} = 12x + \frac{162}{x^2}$ Thus, we have shown that: $L = 12x + \frac{162}{x^2}$

Step 2

Use calculus to find the minimum value of $L$

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Answer

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

$\frac{dL}{dx} = 12 - \frac{324}{x^3}$ Setting the derivative equal to zero to find critical points: $12 - \frac{324}{x^3} = 0 \\ \Rightarrow 12 = \frac{324}{x^3} \\ \Rightarrow x^3 = \frac{324}{12} \\ \Rightarrow x^3 = 27 \\ \Rightarrow x = 3$

Next, we substitute $x = 3$ back into the original equation for $L$ : $L = 12(3) + \frac{162}{3^2} = 36 + \frac{162}{9} = 36 + 18 = 54$ Hence, the minimum value of $L$ is 54 cm.

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, we will differentiate $\frac{dL}{dx}$ to find $\frac{d^2L}{dx^2}$ :

$\frac{d^2L}{dx^2} = \frac{972}{x^4}$ Evaluating the second derivative at $x = 3$ : $\frac{d^2L}{dx^2} = \frac{972}{3^4} = \frac{972}{81} = 12$ Since $\frac{d^2L}{dx^2} > 0$ , we conclude that $L$ has a local minimum at $x = 3$ . Thus, the value of $L$ found 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 1 - 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 1 - 2011 - Paper 2

Show that the total length, $L$ cm, of the twelve edges of the cuboid is given by $L = 12x + \frac{162}{x^2}$

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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