A closed 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.

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Answer:

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

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