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
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
1 answer
2 answers