Aclosed rectangular box is to constructed with a base that is twice as long as it is wide. if the total surface area is 27 square feet, find the dimensions of the box that maximize the volume.

For maximum volume the dimension should be;

w= 1.5ft , b=3ft and h= 2ft

Step-by-step explanation:

The surface area of the rectangular box

S = 2wb + 2wh + 2bh = 27ft^2 1

Given that b = 2w

Substituting into eqn 1

2w(2w) + 2wh + 2h(2w) = 27

4w^2 + 2wh + 4hw = 27

h(6w) = 27 - 4w^2

h = (27 - 4w^2)/6w 2

The volume of a rectangular box is given as

V=w×b×h

V= w×2w×h = 2hw^2 3

Substituting eqn2 into eqn3

V=2w^2(27-4w^2)/6w

V = w(27-4w^2)/3 = (27w-4w^3)/3

To find the maximum point, we need to differentiate the eqn above.

At maximum dV/dw = 0

dV/dw = (27 - 12w^2)/3 = 0

12w^2 = 27

w^2 = 9/4

w = 3/2ft = 1.5ft

b = 2w = 6/2 = 3ft

h = (27 - 4w^2)/6w

h = (27 - 4(3/2)^2)/6(3/2)

h = ( 27 - 9)/6 = 18/9

h = 2ft

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol