3. Find the height and radius of a closed cylinder of volume $$ 125 \mathrm{~cm}^{3} $$ which has the least surface area. Hence calculate the minimum area of the sheet metal required if $$ 25 \% $$ of the material is lost as scrape. 4. A swimming pool is 55 m long and 10 m wide. The perpendicular depth at the deep end is 5 m and at the shallow end is 1.5 m , the slope from one end to the other being uniform. The inside of the pool needs two coats of a protective paint before it is filled with water. Determine how many litres of paint will be needed if 1 litre covers $$ 10 \mathrm{~m}^{2} $$.

Question

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### Problem 3: Closed Cylinder with Minimum Surface Area

**Known Conditions:**
- Volume $$ V = 125 \, \text{cm}^3 $$
- We need to find the height $$ h $$ and radius $$ r $$ of the cylinder that minimizes the surface area.

**Step 1: Formulas**
- Volume of a cylinder: 
  $$
  V = \pi r^2 h
  $$
- Surface area of a closed cylinder:
  $$
  A = 2\pi r^2 + 2\pi rh
  $$

**Step 2: Express $$ h $$ in terms of $$ r $$ using the volume formula:**
$$
h = \frac{V}{\pi r^2} = \frac{125}{\pi r^2}
$$

**Step 3: Substitute $$ h $$ into the surface area formula:**
$$
A = 2\pi r^2 + 2\pi r \left(\frac{125}{\pi r^2}\right)
$$
This simplifies to:
$$
A = 2\pi r^2 + \frac{250}{r}
$$

**Step 4: Differentiate $$ A $$ with respect to $$ r $$ and set the derivative to zero to find the minimum:**
$$
\frac{dA}{dr} = 4\pi r - \frac{250}{r^2}
$$
Setting the derivative to zero:
$$
4\pi r - \frac{250}{r^2} = 0
$$

**Step 5: Solve for $$ r $$:**
$$
4\pi r^3 = 250 \implies r^3 = \frac{250}{4\pi} \implies r = \left(\frac{250}{4\pi}\right)^{1/3}
$$

**Step 6: Calculate $$ r $$ and then find $$ h $$:**
Let's calculate $$ r $$ first.

Now, I will calculate $$ r $$ and then $$ h $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\left(\frac{250}{4\pi }\right)^{\frac{1}{3}}$$
- step1: Reduce the fraction:
  $$\left(\frac{125}{2\pi }\right)^{\frac{1}{3}}$$
- step2: Use the properties of exponents:
  $$\frac{125^{\frac{1}{3}}}{\left(2\pi \right)^{\frac{1}{3}}}$$
- step3: Evaluate the power:
  $$\frac{5}{\left(2\pi \right)^{\frac{1}{3}}}$$
- step4: Evaluate the power:
  $$\frac{5}{2^{\frac{1}{3}}\pi ^{\frac{1}{3}}}$$
- step5: Rewrite the expression:
  $$\frac{5}{\sqrt[3]{2}\times \sqrt[3]{\pi }}$$
- step6: Multiply by the Conjugate:
  $$\frac{5\sqrt[3]{2^{2}}\times \sqrt[3]{\pi ^{2}}}{\sqrt[3]{2}\times \sqrt[3]{\pi }\times \sqrt[3]{2^{2}}\times \sqrt[3]{\pi ^{2}}}$$
- step7: Multiply the numbers:
  $$\frac{5\sqrt[3]{4\pi ^{2}}}{\sqrt[3]{2}\times \sqrt[3]{\pi }\times \sqrt[3]{2^{2}}\times \sqrt[3]{\pi ^{2}}}$$
- step8: Multiply the numbers:
  $$\frac{5\sqrt[3]{4\pi ^{2}}}{2\pi }$$
Calculate or simplify the expression $$ 125/(\pi*((250/(4*\pi))^(1/3))^2) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{125}{\left(\pi \left(\left(\frac{250}{4\pi }\right)^{\frac{1}{3}}\right)^{2}\right)}$$
- step1: Remove the parentheses:
  $$\frac{125}{\pi \left(\left(\frac{250}{4\pi }\right)^{\frac{1}{3}}\right)^{2}}$$
- step2: Reduce the fraction:
  $$\frac{125}{\pi \left(\left(\frac{125}{2\pi }\right)^{\frac{1}{3}}\right)^{2}}$$
- step3: Multiply the exponents:
  $$\frac{125}{\pi \left(\frac{125}{2\pi }\right)^{\frac{1}{3}\times 2}}$$
- step4: Multiply:
  $$\frac{125}{\pi \left(\frac{125}{2\pi }\right)^{\frac{2}{3}}}$$
- step5: Multiply the terms:
  $$\frac{125}{\frac{25\sqrt[3]{2\pi }}{2}}$$
- step6: Multiply by the reciprocal:
  $$125\times \frac{2}{25\sqrt[3]{2\pi }}$$
- step7: Reduce the numbers:
  $$5\times \frac{2}{\sqrt[3]{2\pi }}$$
- step8: Multiply:
  $$\frac{5\times 2}{\sqrt[3]{2\pi }}$$
- step9: Multiply:
  $$\frac{10}{\sqrt[3]{2\pi }}$$
- step10: Multiply by the Conjugate:
  $$\frac{10\sqrt[3]{2^{2}\pi ^{2}}}{\sqrt[3]{2\pi }\times \sqrt[3]{2^{2}\pi ^{2}}}$$
- step11: Multiply the numbers:
  $$\frac{10\sqrt[3]{2^{2}\pi ^{2}}}{2\pi }$$
- step12: Reduce the fraction:
  $$\frac{5\sqrt[3]{2^{2}\pi ^{2}}}{\pi }$$
- step13: Evaluate the power:
  $$\frac{5\sqrt[3]{4\pi ^{2}}}{\pi }$$
From the calculations:

1. The radius $$ r $$ of the cylinder is approximately $$ 2.71 \, \text{cm} $$.
2. The height $$ h $$ of the cylinder is approximately $$ 5.42 \, \text{cm} $$.

**Step 7: Calculate the minimum surface area $$ A $$:**
Now, we can substitute $$ r $$ back into the surface area formula:
$$
A = 2\pi r^2 + \frac{250}{r}
$$

Let's calculate $$ A $$.
Calculate the value by following steps:
- step0: Calculate:
  $$2\pi \times 2.70963^{2}+\frac{250}{2.70963}$$
- step1: Convert the expressions:
  $$2\pi \left(\frac{270963}{100000}\right)^{2}+\frac{250}{2.70963}$$
- step2: Divide the numbers:
  $$2\pi \left(\frac{270963}{100000}\right)^{2}+\frac{25000000}{270963}$$
- step3: Multiply the numbers:
  $$\frac{270963^{2}\pi }{512\times 3125^{2}}+\frac{25000000}{270963}$$
- step4: Reduce fractions to a common denominator:
  $$\frac{270963^{2}\pi \times 270963}{512\times 3125^{2}\times 270963}+\frac{25000000\times 512\times 3125^{2}}{270963\times 512\times 3125^{2}}$$
- step5: Multiply the terms:
  $$\frac{270963^{2}\pi \times 270963}{138733056\times 3125^{2}}+\frac{25000000\times 512\times 3125^{2}}{270963\times 512\times 3125^{2}}$$
- step6: Multiply the terms:
  $$\frac{270963^{2}\pi \times 270963}{138733056\times 3125^{2}}+\frac{25000000\times 512\times 3125^{2}}{138733056\times 3125^{2}}$$
- step7: Transform the expression:
  $$\frac{270963^{2}\pi \times 270963+25000000\times 512\times 3125^{2}}{138733056\times 3125^{2}}$$
- step8: Multiply the terms:
  $$\frac{\pi \times 270963^{3}+25000000\times 512\times 3125^{2}}{138733056\times 3125^{2}}$$
- step9: Multiply the terms:
  $$\frac{\pi \times 270963^{3}+12800000000\times 3125^{2}}{138733056\times 3125^{2}}$$
- step10: Reorder the terms:
  $$\frac{270963^{3}\pi +12800000000\times 3125^{2}}{138733056\times 3125^{2}}$$
The minimum surface area $$ A $$ of the cylinder is approximately $$ 138.40 \, \text{cm}^2 $$.

**Step 8: Calculate the area of the sheet metal required after accounting for scrap:**
If $$ 25\% $$ of the material is lost as scrap, we need to find $$ 75\% $$ of the area:
$$
\text{Area required} = A \times \frac{75}{100} = 138.40 \times 0.75
$$

Let's calculate this.
Calculate the value by following steps:
- step0: Calculate:
  $$138.4\times 0.75$$
- step1: Multiply the numbers:
  $$103.8$$
The area of the sheet metal required after accounting for scrap is approximately $$ 103.80 \, \text{cm}^2 $$.

---

### Problem 4: Swimming Pool Paint Calculation

**Known Conditions:**
- Length of the pool $$ L = 55 \, \text{m} $$
- Width of the pool $$ W = 10 \, \text{m} $$
- Depth at the deep end $$ D_d = 5 \, \text{m} $$
- Depth at the shallow end $$ D_s = 1.5 \, \text{m} $$

**Step 1: Calculate the surface area of the pool.**
The pool has a trapezoidal cross-section. The average depth $$ D $$ can be calculated as:
$$
D = \frac{D_d + D_s}{2} = \frac{5 + 1.5}{2} = 3.25 \, \text{m}
$$

**Step 2: Calculate the surface area of the sides and bottom.**
- Area of the bottom:
$$
A_b = L \times W = 55 \times 10
$$
- Area of the two sides (length):
$$
A_s = 2 \times (L \times D) = 2 \times (55 \times 3.25)
$$
- Area of the two ends (width):
$$
A_e = 2 \times (W \times D) = 2 \times (10 \times 3.25)
$$

**Step 3: Total surface area $$ A_{total} $$:**
$$
A_{total} = A_b + A_s + A_e
$$

Let's calculate $$ A_b $$, $$ A_s $$, and $$ A_e $$ first.
Calculate the value by following steps:
- step0: Calculate:
  $$2\left(10\times 3.25\right)$$
- step1: Remove the parentheses:
  $$2\times 10\times 3.25$$
- step2: Multiply the terms:
  $$20\times 3.25$$
- step3: Multiply the numbers:
  $$65$$
Calculate or simplify the expression $$ 55*10 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$55\times 10$$
- step1: Multiply the numbers:
  $$550$$
Calculate or simplify the expression $$ 2*(55*3.25) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$2\left(55\times 3.25\right)$$
- step1: Remove the parentheses:
  $$2\times 55\times 3.25$$
- step2: Multiply the terms:
  $$110\times 3.25$$
- step3: Multiply the numbers:
  $$357.5$$
The calculated areas are as follows:

1. Area of the bottom $$ A_b $$ is $$ 550 \, \text{m}^2 $$.
2. Area of the two sides $$ A_s $$ is approximately $$ 357.5 \, \text{m}^2 $$.
3. Area of the two ends $$ A_e $$ is $$ 65 \, \text{m}^2 $$.

**Step 4: Calculate the total surface area $$ A_{total} $$:**
$$
A_{total} = A_b + A_s + A_e = 550 + 357.5 + 65
$$

Let's calculate $$ A_{total} $$.
Calculate the value by following steps:
- step0: Calculate:
  $$550+357.5+65$$
- step1: Add the numbers:
  $$972.5$$
The total surface area $$ A_{total} $$ of the pool is approximately $$ 972.5 \, \text{m}^2 $$.

**Step 5: Calculate the amount of paint needed:**
Since the inside of the pool needs two coats of paint, the total area to be painted is:
$$
\text{Total area to paint} = 2 \times A_{total} = 2 \times 972.5
$$

**Step 6: Calculate the number of litres of paint needed:**
Given that $$ 1 \, \text{litre} $$ covers $$ 10 \, \text{m}^2 $$:
$$
\text{Litres of paint} = \frac{\text{Total area to paint}}{10}
$$

Let's calculate the total area to paint and then the litres of paint needed.
Calculate the value by following steps:
- step0: Calculate:
  $$2\times 972.5$$
- step1: Multiply the numbers:
  $$1945$$
Calculate or simplify the expression $$ (2*972.5)/10 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(2\times 972.5\right)}{10}$$
- step1: Remove the parentheses:
  $$\frac{2\times 972.5}{10}$$
- step2: Multiply the numbers:
  $$\frac{1945}{10}$$
- step3: Reduce the fraction:
  $$\frac{389}{2}$$
The calculations yield the following results:

1. The total area to paint is $$ 1945 \, \text{m}^2 $$.
2. The amount of paint needed is approximately $$ 194.5 \, \text{litres} $$.

### Summary of Results:
- **Cylinder Problem:**
  - Radius $$ r \approx 2.71 \, \text{cm} $$
  - Height $$ h \approx 5.42 \, \text{cm} $$
  - Minimum surface area required (after accounting for scrap) $$ \approx 103.80 \, \text{cm}^2 $$

- **Swimming Pool Problem:**
  - Total surface area to paint $$ \approx 972.5 \, \text{m}^2 $$
  - Litres of paint needed $$ \approx 194.5 \, \text{litres} $$
**Cylinder Problem:** - Radius: approximately 2.71 cm - Height: approximately 5.42 cm - Minimum surface area after scrap: approximately 103.80 cm² **Swimming Pool Problem:** - Total surface area to paint: approximately 972.5 m² - Litres of paint needed: approximately 194.5 litres

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Problem 3: Closed Cylinder with Minimum Surface Area

Problem 4: Swimming Pool Paint Calculation

Summary of Results:

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