A-Level Edexcel Maths Pure Proof: Figure 4 shows an open-topped wa

Figure 4 shows an open-topped water tank, in the shape of a cuboid, which is made of sheet metal - Edexcel - A-Level Maths Pure - Question 9 - 2008 - Paper 2

Question 9

Figure 4 shows an open-topped water tank, in the shape of a cuboid, which is made of sheet metal. The base of the tank is a rectangle $x$ metres by $y$ metres. The h... show full transcript

Worked Solution & Example Answer:Figure 4 shows an open-topped water tank, in the shape of a cuboid, which is made of sheet metal - Edexcel - A-Level Maths Pure - Question 9 - 2008 - Paper 2

Step 1

Show that the area $A$ m$^2$ of the sheet metal used to make the tank is given by:

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Answer

To find the surface area of the open-topped tank, we note that it has dimensions $x$ (height) and $y$ (length). The volume of the tank is given by: $V = x \cdot y \cdot x = x^2y$ Since the volume is 100 m $^3$ , we can express $y$ in terms of $x$ : $y = \frac{100}{x^2}.$

The surface area $A$ consists of the base area and the vertical sides: $A = xy + 2xh$ Substituting $h = x$ and $y = \frac{100}{x^2}$ , we have: $A = \frac{100}{x} + 2x^2.$

Thus, we have shown that: $A = \frac{300}{x} + 2x^2.$

Step 2

Use calculus to find the value of $x$ for which $A$ is stationary.

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Answer

We need to find the stationary points of $A$ . To do this, we will differentiate $A$ with respect to $x$ and set it to zero:

$\frac{dA}{dx} = -\frac{300}{x^2} + 4x.$ Setting this equal to zero gives: $-\frac{300}{x^2} + 4x = 0.$ Multiplying through by $x^2$ results in: $-300 + 4x^3 = 0$ Thus,

ightarrow x^3 = 75 ightarrow x = \sqrt[3]{75} \approx 4.22.$$

Step 3

Prove that this value of $x$ gives a minimum value of $A$.

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Answer

To confirm that the value of $x$ we found is a minimum, we can use the second derivative test. We first find the second derivative of $A$ :

$\frac{d^2A}{dx^2} = \frac{600}{x^3} + 4.$ Since $\frac{600}{x^3} + 4 > 0$ for $x > 0$ , this indicates that $A$ is at a minimum when $x = \sqrt[3]{75}.$

Step 4

Calculate the minimum area of sheet metal needed to make the tank.

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Answer

Now we can find the minimum area by substituting $x = \sqrt[3]{75}$ back into the area formula:

$A = \frac{300}{\sqrt[3]{75}} + 2(\sqrt[3]{75})^2.$ Calculating each term and simplifying will give us the minimum area of sheet metal needed to make the tank.

Figure 4 shows an open-topped water tank, in the shape of a cuboid, which is made of sheet metal - Edexcel - A-Level Maths Pure - Question 9 - 2008 - Paper 2

Worked Solution & Example Answer:Figure 4 shows an open-topped water tank, in the shape of a cuboid, which is made of sheet metal - Edexcel - A-Level Maths Pure - Question 9 - 2008 - Paper 2

Show that the area $A$ m$^2$ of the sheet metal used to make the tank is given by:

Use calculus to find the value of $x$ for which $A$ is stationary.

Prove that this value of $x$ gives a minimum value of $A$.

Calculate the minimum area of sheet metal needed to make the tank.

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