Let the size of the square (cross-section) be s.
Then we need to maximize
V=s²(24-2s) with respect to s.
First find the derivative and equate to zero:
dV/ds = 48s-6s²=0
means
s=0 or s=8
s=0 corresponds to a minimum volume and
s=8 corresponds to a maximum volume.
So the maximum volume is given by
8x8x8 inches, as is evident by intuition.
A parcel delivery service a package only of the length plus girth (distance around) does not exceed 24 inches.
A) Find the dimensions of a rectangular box with square ends that satisfies the delivery service's restriction and has a maximum volume. What is the maximum volume?
B) Find the dimensions (radius and height) of a cylindrical container that meets the delivery service's restriction and has a maximum volume. What is the maximum volume?
3 answers
but an 8x8x8 box has length+girth = 8+32 = 40 inches, so it will not work.
We need to optimize s^2(24-4s) since a square has 4 sides.
v = 24s^2 - 4s^3
v' = 48s - 12s^2
v'=0 when s=4
So, a 4x4x8 box has max volume.
Do (B) similarly
We need to optimize s^2(24-4s) since a square has 4 sides.
v = 24s^2 - 4s^3
v' = 48s - 12s^2
v'=0 when s=4
So, a 4x4x8 box has max volume.
Do (B) similarly
Good catch, Steve.
Thank you.
Thank you.