Answer:
a) radius: 16.3 cm
b) area: 4122.7 cm²
c) it is an efficient tub, being within 1.1% of optimum area
Step-by-step explanation:
a)
The volume of the tub is given by the formula ...
V = πr²h
Then the radius is ...
r = √(V/(πh)) = √(20000/(24π)) ≈ 16.3 . . . . cm
The radius is about 16.3 cm.
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b)
The total surface area of a cylinder is ...
A = 2πr(r+h)
A = 2π(16.3 cm)(16.3 +24 cm) = 4122.7 cm²
The surface area is about 4,122.7 cm².
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c)
The "most efficient" cylinder is one that minimizes its surface area for a given volume. Using calculus, it can be shown that cylinder will have its height is equal to its diameter. A good estimate of the optimum dimensions can also be found using a graphing calculator.
For the given cylinder, the height is 24 cm, and the diameter is about 32.6 cm. While that is not the most efficient, a graph of surface area versus height (attached) shows it is only about 1.1% away from the optimum surface area. The most efficient tub would have a height of about 29.4 cm.
((this tub area)/(optimum area) -1) × 100% = (4122.65/4078.82 -1) × 100%
≈ 1.075% . . . . difference of given tub from "most efficient"
While this is not the "most" efficient tub, we would say it is "an" efficient tub.
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Additional comment
In our development above, we have found radius as a function of height, as required by the problem statement. If we're to optimize tub dimensions, it is convenient to define height in terms of radius: h = kr, then find the optimum value of k for a fixed volume.
V = πr²h = kπr³
A = 2πr(r +kr) = 2πr²(1+k) = 2π(V/(kπ))^(2/3)(1 +k)
The derivative with respect to k is ...
A' = (some constant)(-2/3k^(-5/3)(1 +k) +k^(-2/3))
This is zero at the optimum, so we can simplify this to ...
0 = -2/3(1 +k) +k . . . . . multiply by k^(5/3)/(some constant)
2/3 = 1/3k . . . . . . . . collect terms, add 2/3
k = 2 . . . . . . . . . multiply by 3
That is, the height is twice the radius, or equal to the diameter for a cylinder with minimum surface area for a given volume.
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In taking the derivative, we used the power rule and the product rule.
(u^a)' = a(u^(a-1))u'
(uv)' = u'v +uv'
