A box is to contain 225in^3 the height of the box is 5 times the width what dimensions minimize the material needed to build the box

Answer 1

V = LWH but H = 5W and V = 225 so: 225 = LW(5W) so L = 225/(5W²)

To minimize the material in the box means to minimize the surface area. Assuming the box has the 4 sides plus top and bottom, the surface area (SA) is:

SA = 2LW + 2WH + 2LH but H = 5W so:

SA = 2LW + 2W(5W) + 2L(5W) and L = 225/(5W²) so:

SA(W) = 2(225/(5W²)(W) + 10W² + 2(225/(5W²)(5W)

SA(W) = 10W² + 540/W

To minimize, take the derivative and set it equal to zero:

SA(W)' = 20W - 540/W²

20W = 540/W²

W³ = 27

W = 3

H = 15

L = 5