A box with a square base and open top must have a volume of 171500 c m 3 . We wish to find the dimensions of the box that minimize the amount of material used. First, find a formula for the surface area of the box in terms of only x , the length of one side of the square base. [Hint: use the volume formula to express the height of the box in terms of x .] Simplify your formula as much as possible.

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

add 5 subtract 1

step-by-step explanation:

because they could drop even lower..