This is calculus problem.
In an open box you have square base with area a ^ 2 and 4 rectangle with area a * h
where a is length of base and h is height of box
Area of an open box:
A = a ^ 2 + 4 a * h
A = a ^ 2 + 4 a * h = 24 Subtract a ^ 2 to both sides
a ^ 2 + 4 a * h - a ^ 2 = 24 - a ^ 2
4 a * h = 24 - a ^ 2 Divide both sides by 4 a
4 a * h / 4 a = ( 24 - a ^ 2 ) / 4 a
h = ( 24 - a ^ 2 ) / 4 a
Volume of box:
V = a ^ 2 * h
V = a ^ 2 * ( 24 - a ^ 2 ) / 4 a
V = a * a * ( 24 - a ^ 2 ) / 4 a
V = a * ( 24 - a ^ 2 ) / 4
V = ( 24 a - a ^ 3 ) / 4
V = 24 a / 4 - a ^ 3 / 4
V = 6 a - a ^ 3 / 4
V = 6 a - ( 1 / 4 ) a ^ 3
First derivative of V:
dV / da = 6 - ( 1 / 4 ) * 3 a ^ 2 = 6 - ( 3 / 4 ) a ^ 2
dV / da = 6 - ( 3 / 4 ) a ^ 2
A function has extreme point ( maxima or minima ) in point where first derivative = 0
6 - ( 3 / 4 ) a ^ 2 = 0 Add ( 3 / 4 ) a ^ 2
6 - ( 3 / 4 ) a ^ 2 + ( 3 / 4 ) a ^ 2 = 0 + ( 3 / 4 ) a ^ 2
6 = ( 3 / 4 ) a ^ 2 Multiply both sides by 4
6 * 4 = ( 3 / 4 ) a ^ 2 * 4
24 = 3 a ^ 2 Divide both sides by 3
24 / 3 = 3 a ^ 2 / 3
8 = a ^ 2
a ^ 2 = 8
a ^ 2 = 4 * 2
a = ± sqroot ( 4 ) * sqroot ( 2 )
a = ± 2 sqroot ( 2 )
Length of base can't be negative so a = 2 sqroot ( 2 )
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 = - ( 3 / 4 ) * 2 a = - 3 / ( 2 * 2 ) * 2 a = - ( 3 / 2 ) a
For a = 2 sqroot ( 2 ) second derivative =
- ( 3 / 2 ) * 2 * sqroot ( 2 ) =
- 3 sqroot ( 2 ) < 0
So for a = 2 sqroot ( 2 ) function has a maximum.
Now :
h = ( 24 - a ^ 2 ) / 4 a
h = [ 24 - [ 2 * sqroot ( 2 ) ] ^ 2 ] / [ 4 * 2 * sqroot ( 2 ) ]
h = ( 24 - 4 * 2 ) / 8 * sqroot ( 2 )
h = ( 24 - 8 ) / [ 8 * sqroot ( 2 ) ]
h = 16 / [ 8 * sqroot ( 2 ) ]
h = 2 * 8 / [ 8 * sqroot ( 2 ) ]
h = 2 / sqroot ( 2 )
h = sqroot ( 2 ) * sqroot ( 2 ) / sqroot ( 2 )
h = sqroot ( 2 )
Maximum volume:
Vmax = a ^ 2 * h = [ 2 * sqroot ( 2 ) ] ^ 2 * sqroot ( 2 ) = 4 * 2 * sqroot ( 2 ) = 8 sqroot ( 2 )
Vmax = 8 sqroot ( 2 )
for a = 2 sqroot ( 2 ) and h = sqroot ( 2 )
What function should be used to maximize the volume of an open box that has a surface area of 24 and has a square bottom?
Not sure how to do this, a step by step answer would be great. Thanks!!
2 answers
This helps SO much! Thank you!!
1 answer