A storage company designs a rectangular box with an open top that has a volume of 220 in3. Each box has a length that is three times its width. Calculate the minimum surface area of one of these boxes. Round your answer to three decimal places.

191.016 in²

Step-by-step explanation:

Let the volume of the cube be V = lwh where l = length of box, w = width of box and h = height of box.

Now l = 3w.

So, V = lwh = (3w)wh = 3w²h

h = V/3w²

Since the rectangular box is open at the top, its total surface area is A = lw + 2lh + 2wh

substituting l = 3w, we have

= (3w)w + 2h(l + w)

= 3w + 2h(3w + w)

= 3w² + 2h(4w)

= 3w² + 8wh

substituting h = V/3w²

= 3w² + 8w(V/3w²)

A = 3w² + 8V/3w

Differentiating A with respect to w and equating it to zero, we have

dA/dw = d[3w² + 8V/3w]dw

dA/dw = d3w²/dw + d(8V/3w)/dw

dA/dw = 6w - 8V/3w²

dA/dw = 0

6w - 8V/3w² = 0

6w = 8V/3w²

6w³ = 8V/3

w³ = 8V/(3 × 6)

w³ = 4V/9

w = ∛(4V/9)

substituting V = 220 in³, we find the minimum value for the width

w = ∛(4 × 220 in³/9)

w = ∛97.7778

w = 4.607 in

To determine if this is a minimum value of width which gives minimum area, we take the second derivative of A. So,

d²A/dw² = d(6w - 8V/3w²)/dw

= 6 + 16V/3w³

substituting w³ = 4V/9, we have

d²A/dw² = 6 + 16V/3(4V/9)

= 6 + (16 × 9)/(3 × 4)

= 6 + 4 × 3

= 6 + 12

= 18

Since d²A/dw² = 18 > 0 , w = ∛(4V/9) = 4.607 in gives a minimum for the surface area.

So, we calculate this area by substituting w = 4.608 in. So,

A = 3w² + 8V/3w

= 3 × (4.607 in)² + (8 × 220 in³)/(3 × 4.607 in)

= 63.6733 in² + 1,760 in³/13.821 in

= 63.6733 in² + 127.3424 in²

= 191.0157 in²

≅ 191.016 in² to three decimal places.

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