No top?
Area=s^2+ h*s
dA/ds=2s+h=0
s=h/2
but volume=s^2*h
h=32/s^2
s=h/2=16/s^2
s=cubrt16
h=2cubrt16
check: volume=s^2*h=cubrt16 ^2 2cubrt16
volume=32
A box with a square base and no top is to have a volume of 32 ft3. What dimensions use the least amount of material (in other words what dimensions give minimum outside surface area)?
2 answers
let the base be x by x ft and the height be h ft
Volume = x^2h
32 = x^2 h
h = 32/x^2
SA = x^2 + 4xh
= x^2 + 4x(32/x^2)
= x^2 + 128/x
d(SA)/dx = 2x - 128/x^2
= 0 for a max/min of SA
2x = 128/x^2
2x^3= 128
x^3 = 64
x = 4
then h = 32/16 = 2
so the box should have a base of 4ft by 4ft and a height of 2 ft
check:
V = 4x4x2 =32
SA = x^2 + 4xh = 16 + 32 = 48
let x = 3.9
h = 32/3.9^2
SA = 3.9^2 + 128/3.9 = 48.03 > 48
let x = 4.1
SA = 4.1^2 + 128/4.1 = 48.03 > 48
Here is the SA graph as shown by Wolfram,
notice that 48 is the minimum y value for x>0
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+%2B+128%2Fx
Volume = x^2h
32 = x^2 h
h = 32/x^2
SA = x^2 + 4xh
= x^2 + 4x(32/x^2)
= x^2 + 128/x
d(SA)/dx = 2x - 128/x^2
= 0 for a max/min of SA
2x = 128/x^2
2x^3= 128
x^3 = 64
x = 4
then h = 32/16 = 2
so the box should have a base of 4ft by 4ft and a height of 2 ft
check:
V = 4x4x2 =32
SA = x^2 + 4xh = 16 + 32 = 48
let x = 3.9
h = 32/3.9^2
SA = 3.9^2 + 128/3.9 = 48.03 > 48
let x = 4.1
SA = 4.1^2 + 128/4.1 = 48.03 > 48
Here is the SA graph as shown by Wolfram,
notice that 48 is the minimum y value for x>0
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+%2B+128%2Fx