A solid glass cylinder, which is used in an expensive laser amplifier, has a volume of $75 \\pi \, cm^3$. The cost of polishing the surface area of this glass cylinder is £2 per cm² for the curved surface area and £3 per cm² for the circular top and base areas. Given that the radius of the cylinder is $r$ cm, (a) show that the cost of the polishing, £C, is given by $$C = 6 \\pi r^2 + \frac{300 \\pi}{r}$$ (b) Use calculus to find the minimum cost of the polishing, giving your answer to the nearest pound. (c) Justify that the answer that you have obtained in part (b) is a minimum.

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A solid glass cylinder, which is used in an expensive laser amplifier, has a volume of $75 \\pi \, cm^3$ - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 1

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A solid glass cylinder, which is used in an expensive laser amplifier, has a volume of $75 \\pi \, cm^3$. The cost of polishing the surface area of this glass cyli... show full transcript

Worked Solution & Example Answer:A solid glass cylinder, which is used in an expensive laser amplifier, has a volume of $75 \\pi \, cm^3$ - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 1

Step 1

show that the cost of the polishing, £C, is given by

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Answer

To find the cost of polishing the solid glass cylinder, we first need to determine its surface area, which includes the curved surface area and the area of the two circular bases.

Curved Surface Area: The curved surface area of a cylinder is given by:
$A_{curved} = 2 \\pi rh$
Where $h$ is the height.
Area of the Two Circular Bases: The top and bottom circular areas can be calculated as:
$A_{bases} = 2 \\pi r^2$
Expressing Height: From the volume of the cylinder, we know:
$V = 75 \\pi = \\pi r^2 h$
This leads to:
$h = \frac{75}{r^2}$
Substituting Height into the Curved Surface Area: Now by substituting this value of $h$ into the curved surface area formula, we get:
$A_{curved} = 2 \\pi r \left(\frac{75}{r^2}\right) = \frac{150 \\pi}{r}$
Total Surface Area: Therefore, the total surface area $A$ of the cylinder is:
$A_{total} = A_{curved} + A_{bases} = \frac{150 \\pi}{r} + 2 \\pi r^2$
Cost of Polishing: The cost of polishing is given as:
- £2 per cm² for curved surface:
  $Cost_{curved} = 2 * A_{curved} = 2 * \frac{150 \\pi}{r}$
- £3 per cm² for bases:
  $Cost_{bases} = 3 * A_{bases} = 3 * (2 \\pi r^2) = 6 \\pi r^2$

Thus, the total cost C is: $C = 6 \\pi r^2 + \frac{300 \\pi}{r}$

Step 2

Use calculus to find the minimum cost of the polishing, giving your answer to the nearest pound.

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Answer

To find the minimum cost, we need to differentiate the cost function and set the derivative to zero.

Differentiate the Cost Function:
$C = 6 \\pi r^2 + \frac{300 \\pi}{r}$
We differentiate with respect to $r$ :
$\frac{dC}{dr} = 12 \\pi r - \frac{300 \\pi}{r^2}$
Setting the Derivative to Zero:
$12 \\pi r - \frac{300 \\pi}{r^2} = 0$
Simplifying leads to:
$12r^3 = 300 \\Rightarrow r^3 = 25 \\Rightarrow r = 3.21$ (approximately)
Finding the Cost at Minimum Radius:
Substituting $r = 3.21$ back to the cost function: $C = 6 \\pi (3.21)^2 + \frac{300 \\pi}{3.21}$
Compute value to find the minimum cost, rounding it to the nearest pound.

Step 3

Justify that the answer that you have obtained in part (b) is a minimum.

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Answer

To justify that the value calculated in part (b) is indeed a minimum, we can use the second derivative test.

Find the Second Derivative:
Differentiating $\frac{dC}{dr}$ again gives: $\frac{d^2C}{dr^2} = 12 \\pi + \frac{600 \\pi}{r^3}$
Evaluate the Second Derivative at Critical Point:
Substitute $r = 3.21$ :
Since both terms are positive, it implies $\frac{d^2C}{dr^2} > 0$ . Therefore, the function is concave up at this point, confirming a local minimum.

Thus, the answer from part (b) is indeed a minimum cost.

A solid glass cylinder, which is used in an expensive laser amplifier, has a volume of $75 \\pi \, cm^3$ - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 1

Worked Solution & Example Answer:A solid glass cylinder, which is used in an expensive laser amplifier, has a volume of $75 \\pi \, cm^3$ - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 1

show that the cost of the polishing, £C, is given by

Use calculus to find the minimum cost of the polishing, giving your answer to the nearest pound.

Justify that the answer that you have obtained in part (b) is a minimum.

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