Certainly! Let's break down the task step-by-step to understand what is being asked and how to approach. --- ## **Understanding the ProblemA customer wants a **rectangular stainless steel vat** with the following features: - **Shape:** Open-top (no lid) - **Base:** Square - **Volume:** 32 cubic feet (ft³) - **Material:** Stainless steel (for construction, but mainly relevant for surface area calculation) ### **Goals:** 1. **Find the optimal surface area of the vat** (minimize it). 2. **Express the surface area as a function of one variable.** 3. **Determine the minimum surface area** using calculus (by following 5 steps from Section 4.6). --- ## **Step-by-Step Solution** ### **Step 1: Define variables** Let: - \( x \) = length of one side of the square base (feet) - \( h \) = height of the vat (feet) Since the base is square, the area of the base is \( x^2 \). --- ### **Step 2: Write the volume constraint** The volume of the rectangular tank is: \[ \text{Volume} = \text{area of base} \times \text{height} = x^2 \times h \] Given that the volume is 32 ft³: \[ x^2 h = 32 \] Solve for \( h \): \[ h = \frac{32}{x^2} \] --- ### **Step 3: Write the surface area function** The surface area (SA) includes: - The **bottom** (square base): \( x^2 \) - The **four sides**: each has an area of \( x \times h \) Total surface area: \[ S = \text{area of bottom} + 4 \times \text{area of sides} \] \[ S = x^2 + 4(xh) \] Substitute \( h = \frac{32}{x^2} \): \[ S = x^2 + 4 \times x \times \frac{32}{x^2} \] Simplify: \[ S = x^2 + 4 \times \frac{32x}{x^2} \] \[ S = x^2 + \frac{128x}{x^2} \] Since \( \frac{x}{x^2} = \frac{1}{x} \), the expression becomes: \[ S = x^2 + \frac{128}{x} \] --- ### **Step 4: Find the critical points to minimize surface area** To minimize \( S \), take the derivative with respect to \( x \): \[ S' = \frac{d}{dx} \left( x^2 + \frac{128}{x} \right) \] Calculate derivatives: - Derivative of \( x^2 \) is \( 2x \). - Derivative of \( \frac{128}{x} \) is \( -\frac{128}{x^2} \). So, \[ S' = 2x - \frac{128}{x^2} \] Set \( S' = \) to find critical points: \[ 2x - \frac{128}{x^2} = \] Rearranged: \[ 2x = \frac{128}{x^2} \] Multiply both sides by \( x^2 \): \[ 2x \times x^2 = 128 \] \[ 2x^3 = 128 \] Divide both sides by 2: \[ x^3 = 64 \] Take the cube root: \[ x = \sqrt[3]{64} = 4 \] --- ### **Step 5: Find the corresponding height and surface area** Recall: \[ h = \frac{32}{x^2} = \frac{32}{(4)^2} = \frac{32}{16} = 2 \] Calculate the surface area at \( x = 4 \): \[ S = x^2 + \frac{128}{x} = 16 + \frac{128}{4} = 16 + 32 = 48 \] --- ## **Final Answer:** - **Optimal dimensions:** \[ \boxed{ x = 4 \text{ ft (side of square base)}, \quad h = 2 \text{ ft} } \] - **Minimum surface area:** \[ \boxed{ S = 48 \text{ square feet} } \] This configuration minimizes the surface area of the open-top rectangular stainless steel vat given the volume constraint.

## **Understanding the Ant Colony Growth Problem** We need to set up a differential equation to model the growth of an ant colony given a specific growth rate. Let's break it down step-by-step. --- ## **Step 1: Define Variables** Let: - \( P(t) \) = the number of ants in the colony at time \( t \) (measured in months) - \( \frac{dP}{dt} \) = the rate of change of the population with respect to time --- ## **Step 2: Set Up the Differential Equation** According to the problem, the number of ants is increasing at a rate of 600 more than \( \frac{1}{6} \) of the population. This can be expressed mathematically as: \[ \frac{dP}{dt} = \frac{1}{6} P + 600 \] This equation describes how the population changes over time. --- ## **Step 3: Solve the Differential Equation** ### 3.1: Rearranging the Equation First, we can write the equation as: \[ \frac{dP}{dt} - \frac{1}{6} P = 600 \] ### 3.2: Use Integrating Factor The integrating factor \( \mu(t) \) is given by: \[ \mu(t) = e^{\int -\frac{1}{6} dt} = e^{-\frac{1}{6} t} \] ### 3.3: Multiply Through by the Integrating Factor Multiply the entire equation by the integrating factor: \[ e^{-\frac{1}{6} t} \frac{dP}{dt} - \frac{1}{6} e^{-\frac{1}{6} t} P = 600 e^{-\frac{1}{6} t} \] ### 3.4: Recognize the Left Side as a Derivative The left side can be rewritten as the derivative of a product: \[ \frac{d}{dt}(P e^{-\frac{1}{6} t}) = 600 e^{-\frac{1}{6} t} \] ### 3.5: Integrate Both Sides Integrate both sides with respect to \( t \): \[ P e^{-\frac{1}{6} t} = \int 600 e^{-\frac{1}{6} t} dt \] Using integration by substitution, the integral evaluates to: \[ P e^{-\frac{1}{6} t} = -3600 e^{-\frac{1}{6} t} + C \] ### 3.6: Solve for \( P(t) \) Multiply through by \( e^{\frac{1}{6} t} \): \[ P(t) = -3600 + Ce^{\frac{1}{6} t} \] ### 3.7: Initial Condition Given that there are currently 200,000 ants: \[ P(0) = 200000 = -3600 + C \] Solving for \( C \): \[ C = 200000 + 3600 = 203600 \] ### 3.8: Final Expression for \( P(t) \) Thus, the population at time \( t \) is: \[ P(t) = -3600 + 203600 e^{\frac{1}{6} t} \] --- ## **Step 4: Calculate Population After One Year** To find \( P(12) \) (for \( t = 12 \) months): \[ P(12) = -3600 + 203600 e^{\frac{1}{6} \times 12} \] Calculating \( e^{2} \) (approximately 7.389): \[ P(12) = -3600 + 203600 \times 7.389 \] \[ P(12) \approx -3600 + 1505457.4 \approx 1501857.4 \] ### **Rounding to the Nearest Integer:** \[ \boxed{1501857} \text{ ants} \] --- ## **Final Answer** After one year, the estimated population of the ant colony will be approximately **1,501,857 ants**.