Let's solve each part step by step: --- ## (a) **Total Outer Surface Area in Terms of Radius Only** **Given:** - Cylindrical tank with **no top**, **flat circular base**. - **Volume:** \( V = 90\,\text{m}^3 \) - **Let:** radius = \( r \), height = \( h \) **Surface Area (S):** - Side (lateral) area: \( 2\pi r h \) - Base area: \( \pi r^2 \) - **No top.** - **Total surface area:** \[ S = 2\pi r h + \pi r^2 \] **Volume constraint:** \[ V = \pi r^2 h = 90 \implies h = \frac{90}{\pi r^2} \] **Substitute for \( h \):** \[ S = 2\pi r \left(\frac{90}{\pi r^2}\right) + \pi r^2 = \frac{180}{r} + \pi r^2 \] ### **Final Answer (a):** \[ \boxed{S = \frac{180}{r} + \pi r^2} \] where \( S \) is in \( \text{m}^2 \), \( r \) in meters. --- ## (b) **Optimal Dimensions (Minimize Outer Surface Area)** **Given \( S(r) = \frac{180}{r} + \pi r^2 \)** **To minimize:** Find \( r \) such that \( \frac{dS}{dr} = 0 \). ### **Step 1:** Differentiate \( S \) with respect to \( r \): \[ \frac{dS}{dr} = -\frac{180}{r^2} + 2\pi r \] ### **Step 2:** Set derivative to zero for minimum: \[ -\frac{180}{r^2} + 2\pi r = 0 \] \[ 2\pi r = \frac{180}{r^2} \] \[ 2\pi r^3 = 180 \] \[ r^3 = \frac{90}{\pi} \] \[ r = \left(\frac{90}{\pi}\right)^{1/3} \] ### **Step 3:** Find corresponding \( h \): \[ h = \frac{90}{\pi r^2} \] Substitute \( r \): \[ r^2 = \left(\frac{90}{\pi}\right)^{2/3} \] \[ h = \frac{90}{\pi \left(\frac{90}{\pi}\right)^{2/3}} = 90 \cdot \pi^{-1} \cdot \left(\frac{\pi}{90}\right)^{2/3} \] Simplify: \[ h = 90 \cdot \pi^{-1} \cdot \pi^{2/3} \cdot 90^{-2/3} = 90 \cdot \pi^{-1 + 2/3} \cdot 90^{-2/3} = 90 \cdot \pi^{-1/3} \cdot 90^{-2/3} \] \[ = 90^{1-2/3} \cdot \pi^{-1/3} = 90^{1/3} \cdot \pi^{-1/3} = \left(\frac{90}{\pi}\right)^{1/3} \] So, **\( h = r \)**. ### **Step 4:** Second Derivative Test (Relative Minimum) \[ \frac{d^2 S}{dr^2} = \frac{360}{r^3} + 2\pi \] At \( r > 0 \), this is always **positive**, so it is a **minimum**. --- ### **Final Answer (b):** - **Optimal Radius:** \[ r = \left(\frac{90}{\pi}\right)^{1/3} \] - **Optimal Height:** \[ h = r \] - **It is a relative minimum.** --- ## (c) **Maximum Volume for Fixed Surface Area** Let fixed surface area be \( S \). - **From (a):** \( S = 2\pi r h + \pi r^2 \) - **Volume:** \( V = \pi r^2 h \) **Express \( h \) from surface area:** \[ S = 2\pi r h + \pi r^2 \implies 2\pi r h = S - \pi r^2 \implies h = \frac{S - \pi r^2}{2\pi r} \] **Volume:** \[ V = \pi r^2 \left(\frac{S - \pi r^2}{2\pi r}\right) = \frac{r(S - \pi r^2)}{2} \] \[ V = \frac{rS}{2} - \frac{\pi r^3}{2} \] --- ### **Final Answer (c):** \[ \boxed{ V = \frac{rS}{2} - \frac{\pi}{2} r^3 } \] where \( V \) is the maximum volume for a fixed surface area \( S \), in terms of \( S \) and \( r \). --- **If you need numerical answers, plug in values as needed!**

Certainly! Let's carefully go through each part with detailed calculations, explanations, and final answers. --- ## **Part (a): Show that the total outer surface area of the tank (including base and side, but excluding top) can be expressed in terms of radius only.** ### **Given Data:** - The tank is cylindrical with **no top**. - Volume of the tank: \( V = 90\, \text{m}^3 \). - Let: - \( r \) = radius of the base (meters), - \( h \) = height of the cylinder (meters). ### **Step 1: Write down the surface area \( S \)** - The **outer surface area** includes: - **Lateral surface area**: \( 2\pi r h \), - **Base area**: \( \pi r^2 \), - **No top**. \[ \boxed{ S = 2\pi r h + \pi r^2 } \] ### **Step 2: Express \( h \) in terms of \( r \) using volume constraint** \[ V = \pi r^2 h = 90 \] \[ h = \frac{90}{\pi r^2} \] ### **Step 3: Substitute \( h \) into the surface area \( S \)** \[ S = 2\pi r \left(\frac{90}{\pi r^2}\right) + \pi r^2 \] Simplify: \[ S = 2\pi r \times \frac{90}{\pi r^2} + \pi r^2 \] \[ S = \frac{2\pi r \times 90}{\pi r^2} + \pi r^2 \] \[ S = \frac{180}{r} + \pi r^2 \] ### **Final Expression:** \[ \boxed{ \boxed{ S(r) = \frac{180}{r} + \pi r^2 } } \] This expresses the total outer surface area in terms of the radius \( r \) only. --- ## **Part (b): Using partial differentiation, determine the optimal dimensions of the tank that minimize the outer surface area. Prove that the result is a relative minimum.** ### **Step 1: Find the critical points by differentiating \( S(r) \)** \[ S(r) = \frac{180}{r} + \pi r^2 \] Differentiate with respect to \( r \): \[ \frac{dS}{dr} = -\frac{180}{r^2} + 2\pi r \] ### **Step 2: Set derivative to zero to find critical points** \[ -\frac{180}{r^2} + 2\pi r = 0 \] \[ 2\pi r = \frac{180}{r^2} \] Multiply both sides by \( r^2 \): \[ 2\pi r^3 = 180 \] \[ r^3 = \frac{180}{2\pi} = \frac{90}{\pi} \] Thus, \[ \boxed{ r = \left(\frac{90}{\pi}\right)^{1/3} \] ### **Step 3: Find the corresponding height \( h \)** Recall: \[ h = \frac{90}{\pi r^2} \] Calculate \( r^2 \): \[ r^2 = \left(\frac{90}{\pi}\right)^{2/3} \] Substitute into \( h \): \[ h = \frac{90}{\pi \times \left(\frac{90}{\pi}\right)^{2/3}} \] Express denominator: \[ \pi \times \left(\frac{90}{\pi}\right)^{2/3} = \pi \times \frac{(90)^{2/3}}{\pi^{2/3}} = \pi^{1 - 2/3} \times (90)^{2/3} = \pi^{1/3} \times (90)^{2/3} \] Now, \[ h = \frac{90}{\pi^{1/3} \times (90)^{2/3}} \] Express numerator as \( 90 \): \[ h = \frac{90}{\pi^{1/3} \times (90)^{2/3}} \] Rearranged: \[ h = 90 \times \frac{1}{\pi^{1/3} \times (90)^{2/3}} \] Note that: \[ (90)^{1/3} \times (90)^{1/3} = (90)^{2/3} \] So, \[ h = 90 \times \frac{1}{\pi^{1/3} \times (90)^{2/3}} = 90 \times \frac{1}{\pi^{1/3} \times (90)^{2/3}} \] Express 90 as \( (90)^{1/3} \) cubed, so: \[ h = 90 \times \frac{1}{\pi^{1/3} \times (90)^{2/3}} = 90 \times \frac{1}{\pi^{1/3} \times (90)^{2/3}} \] But the key insight is that: \[ h = r \] since earlier calculations show that at the critical point, \( h = r \). --- ### **Step 4: Verify the nature of the critical point** Calculate second derivative: \[ \frac{d^2 S}{dr^2} = \frac{360}{r^3} + 2\pi \] - For \( r > 0 \), both terms are positive: \[ \frac{360}{r^3} > 0, \quad 2\pi > 0 \] - The second derivative is positive, so the critical point corresponds to a **minimum**. --- ### **Final Answer for (b):** \[ \boxed{ \text{Optimal radius}:\quad r = \left(\frac{90}{\pi}\right)^{1/3} } \] \[ \boxed{ \text{Corresponding height}:\quad h = r = \left(\frac{90}{\pi}\right)^{1/3} } \] --- ## **Part (c): If the tank is to be manufactured with a fixed surface area \( S \), express the maximum volume \( V \) achievable in terms of \( S \) and \( r \).** ### **Step 1: Recall the surface area formula** \[ S = 2\pi r h + \pi r^2 \] Solve for \( h \): \[ 2\pi r h = S - \pi r^2 \] \[ h = \frac{S - \pi r^2}{2 \pi r} \] --- ### **Step 2: Write volume \( V \) in terms of \( r \) and \( S \)** \[ V = \pi r^2 h \] Substitute \( h \): \[ V(r) = \pi r^2 \times \frac{S - \pi r^2}{2 \pi r} \] Simplify: \[ V(r) = \frac{\pi r^2}{2 \pi r} (S - \pi r^2) = \frac{r}{2} (S - \pi r^2) \] ### **Step 3: Final expression for \( V \)** \[ \boxed{ V(r) = \frac{r S}{2} - \frac{\pi r^3}{2} } \] This gives the **maximum volume** achievable for a given fixed surface area \( S \), in terms of radius \( r \). --- ## **Summary of Results:** | Part | Result | Explanation | |---|---|---| | (a) | \( S(r) = \frac{180}{r} + \pi r^2 \) | Surface area expressed solely in terms of \( r \). | | (b) | \( r = \left( \frac{90}{\pi} \right)^{1/3} \), \( h = r \) | Dimensions minimizing the surface area; the minimum is confirmed by the positive second derivative. | | (c) | \( V(r) = \frac{r S}{2} - \frac{\pi r^3}{2} \) | Maximum volume for fixed surface area \( S \), as a function of \( r \). | --- **Let me know if you'd like numerical evaluations or further clarifications!**

Sure! Here are the final answers: **(a)** \[ S(r) = \frac{180}{r} + \pi r^2 \] **(b)** Optimal radius: \[ r = \left(\frac{90}{\pi}\right)^{1/3} \] Optimal height: \[ h = r = \left(\frac{90}{\pi}\right)^{1/3} \] **(c)** Maximum volume for fixed surface area \( S \): \[ V(r) = \frac{r S}{2} - \frac{\pi r^3}{2} \]

Certainly! Here are the final answers with brackets: **(a)** \[ S(r) = \frac{180}{r} + \pi r^2 \] **(b)** Optimal radius: \[ r = \left(\frac{90}{\pi}\right)^{1/3} \] Corresponding height: \[ h = r = \left(\frac{90}{\pi}\right)^{1/3} \] **(c)** Maximum volume for fixed surface area \( S \): \[ V(r) = \frac{r S}{2} - \frac{\pi r^3}{2} \]

Certainly! Here are the final answers boxed: **(a)** \[ \boxed{S(r) = \frac{180}{r} + \pi r^2} \] **(b)** \[ \boxed{r = \left(\frac{90}{\pi}\right)^{1/3}} \quad \text{and} \quad h = r = \left(\frac{90}{\pi}\right)^{1/3} \] **(c)** \[ \boxed{V(r) = \frac{r S}{2} - \frac{\pi r^3}{2}} \]