Question

a cylindrical storage container with an open top is to have a volume of ( 16pi mathrm{m}^{3} ). material for the base weighs ( 4 mathrm{~kg} ) per square meter. material for the sides weighs ( 2 mathrm{~kg} ) per square meter. what is the smallest possible weight of such a container? ( 24pi mathrm{kg} ) ( 32pi mathrm{kg} ) ( 40pi mathrm{kg} ) ( 48pi mathrm{kg} ) ( 64pi mathrm{kg} )

Explanation:

Step1: Express volume and surface area

Let the radius of the base be \(r\) and the height be \(h\). The volume \(V=\pi r^{2}h = 16\pi\), so \(h=\frac{16}{r^{2}}\). The surface - area of the base is \(A_{base}=\pi r^{2}\), and the lateral surface - area is \(A_{lateral}=2\pi rh\).
The weight function \(W = 4\pi r^{2}+2\times(2\pi rh)\). Substitute \(h=\frac{16}{r^{2}}\) into the weight function: \(W(r)=4\pi r^{2}+4\pi r\times\frac{16}{r^{2}}=4\pi r^{2}+\frac{64\pi}{r}\), \(r>0\).

Step2: Find the derivative of the weight function

Differentiate \(W(r)\) with respect to \(r\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(W^\prime(r)=8\pi r-\frac{64\pi}{r^{2}}\).
Set \(W^\prime(r) = 0\) to find the critical points:

$$
LATEXBLOCK0
$$

Step3: Use the second - derivative test

Differentiate \(W^\prime(r)\) to get \(W^{\prime\prime}(r)=8\pi+\frac{128\pi}{r^{3}}\).
When \(r = 2\), \(W^{\prime\prime}(2)=8\pi+\frac{128\pi}{8}=8\pi + 16\pi=24\pi>0\). So \(r = 2\) gives a minimum.

Step4: Calculate the minimum weight

Substitute \(r = 2\) into the weight function \(W(r)\).
First, find \(h=\frac{16}{2^{2}} = 4\).
\(W=4\pi\times2^{2}+4\pi\times2\times4\)

$$
LATEXBLOCK1
$$

Answer:

\(48\pi\mathrm{kg}\)