Question

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!!

Answer 1

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 )

Answer 2

This helps SO much! Thank you!!