A rectangular storage container without a lid is to have a volume of 10 m3. The length of its base is twice the width. Material for the base costs $5 per square meter. Material for the sides costs $3 per square meter. Find the cost (in dollars) of materials for the least expensive such container.

$ 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

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step-by-step explanation: