An open rectangular box with square base is to be made from 48 ft^2 of material. What dimensions will result in a box with the largest possible volume?

Are of bottom is : x^2+4

sides 4*xy

Surface area = 48ft^2 = x^2 + 4xy

volume is x^2*y

V = x^2*y

If we have a square base; then:

long=width=x

height = h

Area of this open rectangular box=x²+4 (xh)

then:

x²+4xh=48

4xh=48-x²

h = (48-x²) / 4x

let:

V (x, h) = volume of this open rectangular box

V (x, h) = x²h

V (x) = x² (48-x²) / 4x

V (x) = x (48-x²) / 4

V (x) = (48x-x³) / 4

1) we have to find the first derivative of V (x):

V' (x) = (48-3x²) / 4

2) We find out the values of "x", When V' (x) = 0

(48-3x²) / 4=0

48-3x²=0*4

48-3x²=0

-3x²=-48

x²=-48/-3

x²=16

x=⁺₊√16

We have two possible solutions:

x=-4; this is not solution, because the long or width can't be negative.

x=4; it is possible solutions

3) we have to find the second derivative

V'' (x) = - 6x/4=-3x/2

V'' (4) = - 12/2=-6<0; then we have a maximum at x=4;

4) we find the value of "h", when x=4

h = (48-x²) / 4x

h = (48-4²) / 4 (4)

h = (48-16) / 16

h=32/16

h=2

Therefore:

long=width=4 ft

height=2 ft

V (x, y) = x²h = (4 ft) ² (2ft) = 32 ft³

Answer: the largest possible volume would be 32 ft³