Question
A rectangular block is constructed so that its length is 3 times the width. Find the dimension of the rectangular block that will give the least possible surface area if the volume of the block is 81 cubic meters.
Answers
The block's dimensions are w, 3w, h
v = 3w^2h = 81, so h = 81/(3w^2) = 27/w^2
The area is
a = 2(w*3w + wh + 3wh) = 6w^2 + 4wh = 6w^2 + 4w*27/w^2 = 6w^2 + 108/w
for minimum area, you want da/dw = 0, so
da/dw = 12w - 108/w^2 = 12(w^3-9)/w^2
da/dw=0 when w^3-9 = 0
w = ∛9
...
v = 3w^2h = 81, so h = 81/(3w^2) = 27/w^2
The area is
a = 2(w*3w + wh + 3wh) = 6w^2 + 4wh = 6w^2 + 4w*27/w^2 = 6w^2 + 108/w
for minimum area, you want da/dw = 0, so
da/dw = 12w - 108/w^2 = 12(w^3-9)/w^2
da/dw=0 when w^3-9 = 0
w = ∛9
...
a square cross-section gives the least surface area
so the volume is ... 81 = 3 w * w * w = 3 w^3
the block is 3 x 3 x 9
to Steve ... you lost a factor of two in your area calculation ... a = ....
so the volume is ... 81 = 3 w * w * w = 3 w^3
the block is 3 x 3 x 9
to Steve ... you lost a factor of two in your area calculation ... a = ....
rats. Way to watch.
Im so confused with these. Can anyone elaborate more on this?