Questions: Problem 14.1: (7 points) A manufacturing company wants to construct a rectangular box with a square base and an open top as shown below. Suppose the box needs to have a volume of 108 cm^3. Follow the steps below to find the dimensions of the box that will minimize the surface area (and therefore minimize the cost of materials to manufacture the box). Find the dimensions that will minimize the surface area. To get full credit, you must verify your value is a local minimum with either the First or Second Derivative test. Show all steps to receive credit.

Problem 14.1: (7 points)
A manufacturing company wants to construct a rectangular box with a square base and an open top as shown below. Suppose the box needs to have a volume of 108 cm^3. Follow the steps below to find the dimensions of the box that will minimize the surface area (and therefore minimize the cost of materials to manufacture the box).

Find the dimensions that will minimize the surface area. To get full credit, you must verify your value is a local minimum with either the First or Second Derivative test.

Show all steps to receive credit.

Transcript text: Problem 14.1: (7 points) A manufacturing company wants to construct a rectangular box with a square base and an open top as shown below. Suppose the box needs to have a volume of $108 \mathrm{~cm}^{3}$. Follow the steps below to find the dimensions of the box that will minimize the surface area (and therefore minimize the cost of materials to manufacture the box). Find the dimensions that will minimize the surface area. To get full credit, you must verify your value is a local minimum with either the First or Second Derivative test. Show all steps to receive credit.

Solution

Solution Steps

Step 1: Define the variables and given information

Let $ x $ be the side length of the square base and $ y $ be the height of the box. The volume of the box is given as 108 cm³.

Step 2: Write the volume equation

The volume $ V $ of the box is given by: \[ V = x^2 y \] Given $ V = 108 $ cm³, we have: \[ x^2 y = 108 \] \[ y = \frac{108}{x^2} \]

Step 3: Write the surface area equation

The surface area $ A $ of the box (with an open top) is given by: \[ A = x^2 + 4xy \] Substitute $ y $ from the volume equation: \[ A = x^2 + 4x \left( \frac{108}{x^2} \right) \] \[ A = x^2 + \frac{432}{x} \]

Step 4: Find the critical points

To minimize the surface area, take the derivative of $ A $ with respect to $ x $ and set it to zero: \[ \frac{dA}{dx} = 2x - \frac{432}{x^2} \] Set the derivative equal to zero: \[ 2x - \frac{432}{x^2} = 0 \] \[ 2x = \frac{432}{x^2} \] \[ 2x^3 = 432 \] \[ x^3 = 216 \] \[ x = 6 \]

Step 5: Verify the critical point using the second derivative test

Find the second derivative of $ A $: \[ \frac{d^2A}{dx^2} = 2 + \frac{864}{x^3} \] Substitute $ x = 6 $: \[ \frac{d^2A}{dx^2} \bigg|_{x=6} = 2 + \frac{864}{6^3} \] \[ \frac{d^2A}{dx^2} \bigg|_{x=6} = 2 + \frac{864}{216} \] \[ \frac{d^2A}{dx^2} \bigg|_{x=6} = 2 + 4 \] \[ \frac{d^2A}{dx^2} \bigg|_{x=6} = 6 \] Since the second derivative is positive, $ x = 6 $ is a local minimum.