Answer:
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.
