st \( 1(4.2,4.5-4.9) \quad \) Test Score: \( 0 \%, 0 \) of 113 points A rectangula tank with a square base, an open top, and a volume of \( 16,384 \mathrm{ft}^{3} \) is to be constructed of sheet steel. Find the dimensions of the tank that has the minimum surface area 0 of 8

To find the dimensions of the rectangular tank with a square base, an open top, and a volume of \(16,384 \ \text{ft}^3\) that minimizes the surface area, follow these steps: ### 1. Define Variables - Let \( x \) be the length of one side of the square base (in feet). - Let \( h \) be the height of the tank (in feet). ### 2. Volume Constraint The volume \( V \) of the tank is given by: \[ V = x^2 \cdot h = 16,384 \ \text{ft}^3 \] Solving for \( h \): \[ h = \frac{16,384}{x^2} \] ### 3. Surface Area to Minimize Since the tank has an open top, the surface area \( S \) consists of the base and the four sides: \[ S = x^2 + 4x \cdot h \] Substitute \( h \) from the volume constraint: \[ S = x^2 + 4x \left( \frac{16,384}{x^2} \right) = x^2 + \frac{65,536}{x} \] ### 4. Optimization To find the value of \( x \) that minimizes \( S \), take the derivative of \( S \) with respect to \( x \) and set it to zero: \[ \frac{dS}{dx} = 2x - \frac{65,536}{x^2} = 0 \] Solving for \( x \): \[ 2x = \frac{65,536}{x^2} \\ 2x^3 = 65,536 \\ x^3 = 32,768 \\ x = \sqrt[3]{32,768} \\ x = 32 \ \text{ft} \] ### 5. Find \( h \) Using the volume constraint: \[ h = \frac{16,384}{32^2} = \frac{16,384}{1,024} = 16 \ \text{ft} \] ### 6. Conclusion The dimensions that minimize the surface area are: - **Base:** \( 32 \ \text{ft} \times 32 \ \text{ft} \) - **Height:** \( 16 \ \text{ft} \) These dimensions ensure the tank has the minimum possible surface area while maintaining the required volume. **Answer:** A square base 32 ft on a side and height 16 ft. • Base: 32 ft × 32 ft • Height: 16 ft