Questions: An closed box with a square base is to have a volume of 2900 cm^3. What should the dimensions of the box be if the amount of material used is to be minimum? (Use decimal notation. Give your answers to three decimal places.) Recall: The surface area of a rectangular box is: SA=2 * l * w+2 * l * h+2 * w * h, where l= length, w= width, and h= height. The volume of a box is: V=l * w * h. square base side length: cm height: cm

An closed box with a square base is to have a volume of 2900 cm^3. What should the dimensions of the box be if the amount of material used is to be minimum?
(Use decimal notation. Give your answers to three decimal places.)
Recall: The surface area of a rectangular box is: SA=2 * l * w+2 * l * h+2 * w * h, where l= length, w= width, and h= height. The volume of a box is: V=l * w * h.
square base side length: cm
height: cm

Transcript text: An closed box with a square base is to have a volume of $2900 \mathrm{~cm}^{3}$. What should the dimensions of the box be if the amount of material used is to be minimum? (Use decimal notation. Give your answers to three decimal places.) Recall: The surface area of a rectangular box is: $S A=2 \cdot l \cdot w+2 \cdot l \cdot h+2 \cdot w \cdot h$, where $l=$ length, $w=$ width, and $h=$ height. The volume of a box is: $V=1 \cdot w \cdot h$. square base side length: $\square$ cm height: $\square$ cm

Solution

Solution Steps

To minimize the amount of material used for a closed box with a square base and a given volume, we need to minimize the surface area while maintaining the volume constraint. The box has a square base, so the length and width are equal. Let's denote the side length of the base as $ x $ and the height as $ h $. The volume constraint is $ x^2 \cdot h = 2900 $. The surface area to minimize is $ SA = 2x^2 + 4xh $. We can express $ h $ in terms of $ x $ using the volume constraint and substitute it into the surface area formula. Then, we find the derivative of the surface area with respect to $ x $, set it to zero to find critical points, and determine the dimensions that minimize the surface area.

Step 1: Define the Variables

Let $ x $ be the side length of the square base of the box, and $ h $ be the height of the box. The volume of the box is given by the equation: \[ V = x^2 \cdot h = 2900 \] From this, we can express the height $ h $ in terms of $ x $: \[ h = \frac{2900}{x^2} \]

Step 2: Surface Area Expression

The surface area $ SA $ of the box is given by: \[ SA = 2x^2 + 4xh \] Substituting the expression for $ h $: \[ SA = 2x^2 + 4x \left(\frac{2900}{x^2}\right) = 2x^2 + \frac{11600}{x} \]

Step 3: Find the Derivative

To minimize the surface area, we take the derivative of $ SA $ with respect to $ x $: \[ \frac{d(SA)}{dx} = 4x - \frac{11600}{x^2} \] Setting the derivative equal to zero to find critical points: \[ 4x - \frac{11600}{x^2} = 0 \]

Step 4: Solve for Critical Points

Rearranging gives: \[ 4x^3 = 11600 \implies x^3 = 2900 \implies x = 2900^{1/3} \] The critical points are: \[ x = 2900^{1/3}, \quad x = -\frac{2900^{1/3}}{2} - \frac{2900^{1/3}\sqrt{3}}{2}i, \quad x = -\frac{2900^{1/3}}{2} + \frac{2900^{1/3}\sqrt{3}}{2}i \] Since $ x $ must be a positive real number, we take: \[ x = 2900^{1/3} \]

Step 5: Calculate the Height

Now, substituting $ x $ back into the volume equation to find $ h $: \[ h = \frac{2900}{(2900^{1/3})^2} = 2900^{1/3} \]