Easy calc problem help (10 pts for best answer)!!?

Let's call the width of the box x and the height of the box y.

The volume of the box is:

V = x² y

The area of the material is:

A = 4xy + x²

1600 = 4xy + x²

The goal is to write the volume V in terms of just one variable. To do that, solve for y:

1600 − x² = 4xy

y = (1600 − x²) / 4x

Substitute into the volume equation:

V = x² y

V = (1600 x − x³) / 4

V = 400 x − 1/4 x³

To maximize, find dV/dx and set to 0:

dV/dx = 400 − 3/4 x²

0 = 400 − 3/4 x²

3/4 x² = 400

x² = 1600 / 3

x = 40 / √3

x ≈ 23.1 cm

y ≈ 11.5 cm

V ≈ 6158.4 cm³

Let the base side = b and height = h

Then surface area of base = b²

the four sides surface area = 4 bh

Total surface area = b² + 4bh

1600 = b² + 4bh

4bh = 1600 - b²

bh = (1600 - b²)/4 .....(1)

Volume of the box,

V = b x b x h

plug bh from (1)

V = ¼b(1600 - b²)

V = 400b - ¼b³ ....(2)

To find optimal volume, find derivative with respect to 'b' and equate to 0

V' = 400 - ¾b²

0 = 400 - ¾b²

¾b² = 400

b² = 1600/3

b = 23 cm

plug in (2)

V = 400(23) - ¼(23)³ = 6158.25 cm³

square base; let each side be 'b'

height of box be 'h'

SA = b^2 + 4bh

V = b^2 h

1,600 = b^2 + 4bh

h = (1,600 - b^2) / 4b

V = b^2 [(1,600 - b^2) / 4b]

V = b/4 (1,600 - b^2)

V = 400b - b^3 / 4

dV/db = 0 = 400 - (3/4)b^2

b = 23.1 cm

h = 11.5 cm

V = 6,137 cm^3