Question

An open box with a square base is to have a volume of 12ft^3.

Find the box dimensions that minimize the amount of material used. (round to two decimal places).
it asks for the side length and the height.

Please help asap due in a few hours.
Thank you

Answer 1

In an open box you have square base with area a ^ 2 and 4 rectangle with area a * h

where:

a = length of base

h = height of box


Volume of box:

V = a ^ 2 * h = 12 ft ^ 3

a ^ 2 * h = 12 Divide both sides by a ^ 2


h = 12 / a ^ 2


Area of an open box:

A = a ^ 2 + 4 a * h

A = a ^ 2 + 4 a * 12 / a ^ 2

A = a ^ 2 + 48 / a


First derivative of A:

dA / da = d ( a ^ 2 ) / da + 48 d ( 1 / a ) / da

dA / da = 2 a + 48 * d ( a ^ - 1 ) / da

dA / da = 2 a + 48 * ( - 1 ) * a ^ ( - 1 - 1 )

dA / da = 2 a - 48 * a ^ ( - 2 )

dA / da = 2 a - 48 / a ^ 2


A function has extreme point ( maximum or minimum ) in point where first derivative = 0

2 a - 48 / a ^ 2 = 0 Multiply both sides by a ^ 2

2 a ^ 3 - 48 = 0 Divide both sides by 2

a ^ 3 - 24 = 0 Add 24 to both sides

a ^ 3 - 24 + 24 = 0 + 24

a ^ 3 = 24

a ^ 3 = 8 * 3

a = third root ( 8 * 3 )

a = third root ( 8 ) * third root ( 3 )

a = 2 * third root ( 3 )


Now you must do second derivative test.

If second derivative < 0 function has a maximum.

If second derivative > 0 function has a minimum.


Second derivative = derivative of first derivative =

2 * d ( a ) / da - 48 * d ( 1 / a ^ 2 ) / da =

2 * d ( a ) / da - 48 * d ( a ^ - 2 ) / da =

2 * 1 - 48 * ( - 2 ) * a ^ ( - 2 - 1 ) =

2 + 96 * a ^ ( - 3 ) =

2 + 96 / a ^ 3


For a = 2 * third root ( 3 ) second derivative =

2 + 96 / [ 2 third root ( 3 ) ] ^ 3 =

2 + 96 / [ 2 ^ 3 * third root ( 3 ) ^ 3 ] =

2 + 96 / ( 8 * 3 ) =

2 + 96 / 24 =

2 + 4 = 6 > 0


So for a = 2 * third root ( 3 ) function has a mimum.

a = 2 * third root ( 3 ) = 2 * 1.44224957 = 2.88449914


h = 12 / a ^ 2 =

12 / ( 2.88449914 ) ^ 2 =

12 / 8.3203352886607396 = 1.442249571


a = 2.88 ft

h = 1.44 ft

rounded to two decimal places