Answer:
The dimensions of the box that minimize the amount of material used is length of square base x = 70 cm and height of box = 35 cm
Step-by-step explanation:
The volume of the box with square base is V = x²h where x = length of side of square base and h = height of box.
Now, the area of the of box open at the top with dimensions of length, l , width, w and height, h are
A = 2lh + lw + 2wh
Now for a box of square base, l = w = x.
So, A = 2xh + x² + 2xh
A = x² + 4xh
Since V = x²h and V = 171500 cm³,
h = V/x² = 171500/x²
substituting h into A, we have
A = x² + 4xh
A = x² + 4x(171500 cm³/x²)
A = x² + 4 × 171500 cm³/x
A = x² + 686000 cm³/x
To find the value of x that minimizes A, we differentiate A with respect to x
So, dA/dx = d[x² + 686000 cm³/x]/dx
dA/dx = 2x - 686000 cm³/x²
equating this to zero, we have
2x - 686000 cm³/x² = 0
2x = 686000 cm³/x²
2x³ = 686000
x³ = 686000 cm³/2
x = ∛(343000 cm³)
x = 70 cm
To determine if this in a minimum dimension for A, we differentiate dA/dx and insert x = 70 cm
So, d²A/dx² = d[2x - 686000 cm³/x²]/dx
d²A/dx² = 2 + 3 × 686000 cm³/x³
d²A/dx² = 2 + 2058000 cm³/x³
with x = 70 cm, we have
d²A/dx² = 2 + 2058000 cm³/(70 cm)³
d²A/dx² = 2 + 2058000 cm³/343000 cm³
d²A/dx² = 2 + 6
d²A/dx² = 8 > 0
So, x = 70 is a minimum value for the dimension
Since h = 171500 cm³/x²
h = 171500 cm³/(70 cm)²
h = 171500 cm³/4900 cm²
h = 35 cm
So, the dimensions of the box that minimize the amount of material used is length of square base x = 70 cm and height of box = 35 cm
