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A solid glass cylinder, which is used in an expensive laser amplifier, has a volume of $75 imes ext{π} ext{ cm}^3$ - Edexcel - A-Level Maths Pure - Question 4 - 2015 - Paper 2

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

Worked Solution & Example Answer:A solid glass cylinder, which is used in an expensive laser amplifier, has a volume of $75 imes ext{π} ext{ cm}^3$ - Edexcel - A-Level Maths Pure - Question 4 - 2015 - Paper 2

Step 1

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

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Answer

To determine the cost of polishing, we need to calculate both the curved surface area and the area of the circular top and bottom of the cylinder.

  1. Curved Surface Area: The formula for the curved surface area of a cylinder is: Acurved=2extπrhA_{curved} = 2 ext{π}rh where hh is the height of the cylinder. We can express the height in terms of the radius and volume. Given the volume, V=extπr2hV = ext{π}r^{2}h we can solve for hh: h=75extπr2h = \frac{75}{ ext{π}r^{2}} Thus, Acurved=2extπr(75extπr2)=150rA_{curved} = 2 ext{π}r \left(\frac{75}{ ext{π}r^{2}}\right) = \frac{150}{r}

  2. Area of Top and Bottom: The area for both circular bases is: Atop+bottom=2imesextπr2=2extπr2A_{top+bottom} = 2 imes ext{π}r^{2} = 2 ext{π}r^{2}

  3. Total Surface Area: Therefore, the total surface area that needs polishing is: Atotal=Acurved+Atop+bottom=150r+2extπr2A_{total} = A_{curved} + A_{top+bottom} = \frac{150}{r} + 2 ext{π}r^{2}

  4. Cost Calculation: The cost for polishing:

    • Curved area costs £2 per cm², and top/bottom areas cost £3 per cm²: C=2⋅150r+3(2extπr2)=300r+6extπr2C = 2 \cdot \frac{150}{r} + 3(2 ext{π}r^{2}) = \frac{300}{r} + 6 ext{π}r^{2} Thus, the expression is: C=6extπr2+300extπrC = 6 ext{π}r^{2} + \frac{300 ext{π}}{r}

Step 2

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

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Answer

We need to minimize the cost function: C=6extπr2+300extπrC = 6 ext{π}r^{2} + \frac{300 ext{π}}{r}

  1. Find the derivative: Taking the derivative of CC with respect to rr: C′=12extπr−300extπr2C' = 12 ext{π}r - \frac{300 ext{π}}{r^{2}}

  2. Setting the derivative to zero: 12extπr−300extπr2=012 ext{π}r - \frac{300 ext{π}}{r^{2}} = 0 Multiplying through by r2r^{2} gives: 12extπr3−300extπ=012 ext{π}r^{3} - 300 ext{π} = 0 or 12r3=30012r^{3} = 300 Simplifying,

ightarrow r = 3.14$$ (approximately)

  1. Finding the minimum cost: Substitute r=3.14r = 3.14 back into the cost function: C=6extπ(3.14)2+300extπ3.14C = 6 ext{π}(3.14)^{2} + \frac{300 ext{π}}{3.14} Calculate to find:

ightarrow ext{round to the nearest pound}$$

Step 3

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

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Answer

To verify that the critical point we found is indeed a minimum, we can use the second derivative test:

  1. Find the second derivative: C′′=12extπ+600extπr3C'' = 12 ext{π} + \frac{600 ext{π}}{r^{3}}

  2. Evaluate the second derivative at r=3.14r = 3.14: Since both terms are positive, C′′>0extwhenr=3.14C'' > 0 ext{ when } r = 3.14 which indicates that the cost function is concave up at this point. Therefore, the critical point corresponds to a local minimum.

This justifies that the cost obtained in part (b) is indeed a minimum.

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