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Question: 2) A supermarket employee wants to construct an open-top box from a 16 by 30 in piece of cardboard…
- A supermarket employee wants to construct an open-top box from a 16 by 30 in piece of cardboard. To do this, the employee plans to cut out squares of equal size from the four corners so the four sides can be bent upwards. What size should the squares be in order to create a box with the largest possible volume?
Solution
To solve this problem, we need to find the size of the square cutouts to maximize the volume of the box. Let’s denote the side length of the squares cut out from each corner as \( x \). The original piece of cardboard is 16 by 30. After cutting out squares of side length \( x \) from each corner, the dimensions of the base of the box will be: \[ \begin{cases} \text{Length: } (30 - 2x) & \\ \text{Width: } (16 - 2x) & \end{cases} \] The height of the box will be \( x \). The volume \( V \) of the box is given by: \[ V = (\text{Length}) \times (\text{Width}) \times (\text{Height}) = (30 - 2x)(16 - 2x)x \] Expanding this, we have: \[ V = (30 - 2x)(16 - 2x)x = (480 - 60x - 32x + 4x^2)x \] \[ V = 480x - 92x^2 + 4x^3 \] To maximize the volume, we take the derivative of \( V \) with respect to \( x \) and set it to zero: \[ V’(x) = 480 - 184x + 12x^2 \] Set the derivative equal to zero: \[ 480 - 184x + 12x^2 = 0 \] This is a quadratic equation. Solving for \( x \), we use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 12 \), \( b = -184 \), and \( c = 480 \). \[ x = \frac{184 \pm \sqrt{(-184)^2 - 4 \times 12 \times 480}}{2 \times 12} \] \[ x = \frac{184 \pm \sqrt{33856 - 23040}}{24} \] \[ x = \frac{184 \pm \sqrt{10816}}{24} \] \[ x = \frac{184 \pm 104}{24} \] This gives: \[ x = \frac{288}{24} = 12 \quad \text{or} \quad x = \frac{80}{24} \approx 3.33 \] Since \( x = 12 \) would eliminate the base (as \( 16 - 2 \times 12 < 0 \)), we discard it. Thus, \( x \approx 3.33 \) is valid. The size of the squares should be approximately 3.33 inches to create the box with the largest possible volume.