Answer:
$ 81.77
Step-by-step explanation:
Let the volume of the rectangular storage cylinder be V = lwh where l = length, w = width and h = height of container.
Now, since the container is open at the top, its bottom area is lw and its side area is 2lh + 2wh = 2(l + w)h
So, the cost for the base material is 5lw and that for the side material is 3 × 2(lh + wh) = 6(lh + wh).
The total cost C is thus C = 5lw + 6(l + w)h
We know that the length is twice the width, so. l = 2w and the volume = 10 m³. So, V = lwh = (2w)wh = 2w²h. Thus h = V/2w² = 10/2w² = 5/w².
Substituting l and h into C, we have
C = 5lw + 6(l + w)h
C = 5(2w)w + 6((2w) + w)(5/w²)
C = 10w² + 6(3w)(5/w²)
C = 10w² + 90/w
Now C is a function of w the width
C(w) = 10w² + 90/w
We differentiate C(w) and equate it to zero to find the turning point of C(w)
dC(w)/dt = d{10w² + 90/w)/dt
dC(w)/dt = 20w - 90/w²
equating it to zero, we have
dC(w)/dt = 0
20w - 90/w² = 0
20w = 90/w²
w³ = 90/20
w³ = 4.5
w = ∛4.5
We differentiate dC(w)/dt again and insert w to determine if this is a minum or maximum point. So,
d²C(w)/dt² = d(20w - 90/w²)/dt
d²C(w)/dt² = 20 + 180/w³
substituting w³ = 4.5, we have
d²C(w)/dt² = 20 + 180/4.5
d²C(w)/dt² = 20 + 40 = 60
Since d²C(w)/dt² = 60 > 0
w = ∛4.5 is a minimum for C(w).
So, substituting w = ∛4.5 into C(w), we have
C(w) = 10w² + 90/w
C(w) = 10(∛4.5)² + 90/∛4.5
C(w) = 10(1.65)² + 90/1.65
C(w) = 27.225 + 54.545
C(w) = 81.77
So the cost for the least expensive of such container is $ 81.77
