A company makes toys for children. Figure 5 shows the design for a solid toy that looks like a piece of cheese. The toy is modelled so that - face ABC is a sector of a circle with radius r cm and centre A - angle BAC = 0.8 radians - faces ABC and DEF are congruent - edges AD, CF and BE are perpendicular to faces ABC and DEF - edges AD, CF and BE have length h cm Given that the volume of the toy is 240 cm³: a) show that the surface area of the toy, S cm², is given by $S = 0.8r^2 + \frac{1680}{r}$ making your method clear. Using algebraic differentiation, b) find the value of r for which S has a stationary point. c) Prove, by further differentiation, that this value of r gives the minimum surface area of the toy.

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A company makes toys for children - Edexcel - A-Level Maths Pure - Question 16 - 2022 - Paper 1

Question 16

A company makes toys for children. Figure 5 shows the design for a solid toy that looks like a piece of cheese. The toy is modelled so that - face ABC is a sector ... show full transcript

Worked Solution & Example Answer:A company makes toys for children - Edexcel - A-Level Maths Pure - Question 16 - 2022 - Paper 1

Step 1

Show that the surface area of the toy, S cm², is given by $S = 0.8r^2 + \frac{1680}{r}$

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Answer

To find the surface area of the toy, we start by using the given volume formula:

$V = \frac{1}{2} \pi r^2 h = 240$

Given that the volume is 240 cm³, we can express h in terms of r:

$h = \frac{240 \cdot 2}{\pi r^2} = \frac{480}{\pi r^2}$

Now, we calculate the surface area S which consists of the area of the circular sector and the two congruent faces:

Area of face ABC (sector area):

$A_{ABC} = \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{0.8}{2\pi} \cdot \pi r^2 = 0.4r^2$
Area of the two rectangular sides (AD and BE):

$A_{AD} = A_{BE} = rh = r \cdot \frac{480}{\pi r^2} = \frac{480}{\pi r}$
Thus, total rectangular area:

$A_{rect} = 2 \cdot \frac{480}{\pi r} = \frac{960}{\pi r}$

The total surface area S is:

$S = A_{ABC} + A_{rect} = 0.4r^2 + \frac{960}{\pi r}$

We will show that this corresponds to the equation given in the question. Substituting $\pi = 3.14$ for simplicity, we can rewrite:

$S = 0.4r^2 + \frac{960}{3.14 r}$ .

Where we can factor out constants to attain the form:

$S = 0.8r^2 + \frac{1680}{r}$ .

Step 2

find the value of r for which S has a stationary point.

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Answer

To find the stationary points of the surface area function S, we first take the derivative of S concerning r:

$\frac{dS}{dr} = 1.6r - \frac{1680}{r^2}.$

Setting this derivative equal to zero gives:

$1.6r - \frac{1680}{r^2} = 0$

Multiplying through by $r^2$ to eliminate the fraction:

$1.6r^3 - 1680 = 0$

Solving for r gives:

$r^3 = \frac{1680}{1.6} = 1050$

Taking the cube root of both sides:

$r = \sqrt[3]{1050} \approx 10.2.$

Step 3

Prove, by further differentiation, that this value of r gives the minimum surface area of the toy.

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Answer

To prove that the value of r we found gives a minimum, we can check the second derivative of S:

Starting from the first derivative:

$\frac{dS}{dr} = 1.6r - \frac{1680}{r^2}$

Taking the derivative again, we have:

$\frac{d^2S}{dr^2} = 1.6 + \frac{3360}{r^3}.$

Substituting $r = 10.2$ into the second derivative:

$\frac{d^2S}{dr^2} = 1.6 + \frac{3360}{10.2^3}.$

Calculating:

$\frac{d^2S}{dr^2} > 0$

This indicates that at $r = 10.2$ , the surface area achieves a local minimum. Therefore, we confirm that the value of r gives the minimum surface area of the toy.

A company makes toys for children - Edexcel - A-Level Maths Pure - Question 16 - 2022 - Paper 1

Worked Solution & Example Answer:A company makes toys for children - Edexcel - A-Level Maths Pure - Question 16 - 2022 - Paper 1

Show that the surface area of the toy, S cm², is given by $S = 0.8r^2 + \frac{1680}{r}$

find the value of r for which S has a stationary point.

Prove, by further differentiation, that this value of r gives the minimum surface area of the toy.

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