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
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
Answers
Bosnian
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
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