Respuesta :
Answer:
The dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.
Step-by-step explanation:
A cylindrical can holds 300 cubic centimeters, and we want to find the dimensions that minimize the cost for materials: that is, the dimensions that minimize the surface area.
Recall that the volume for a cylinder is given by:
[tex]\displaystyle V = \pi r^2h[/tex]
Substitute:
[tex]\displaystyle (300) = \pi r^2 h[/tex]
Solve for h:
[tex]\displaystyle \frac{300}{\pi r^2} = h[/tex]
Recall that the surface area of a cylinder is given by:
[tex]\displaystyle A = 2\pi r^2 + 2\pi rh[/tex]
We want to minimize this equation. To do so, we can find its critical points, since extrema (minima and maxima) occur at critical points.
First, substitute for h.
[tex]\displaystyle \begin{aligned} A &= 2\pi r^2 + 2\pi r\left(\frac{300}{\pi r^2}\right) \\ \\ &=2\pi r^2 + \frac{600}{ r} \end{aligned}[/tex]
Find its derivative:
[tex]\displaystyle A' = 4\pi r - \frac{600}{r^2}[/tex]
Solve for its zero(s):
[tex]\displaystyle \begin{aligned} (0) &= 4\pi r - \frac{600}{r^2} \\ \\ 4\pi r - \frac{600}{r^2} &= 0 \\ \\ 4\pi r^3 - 600 &= 0 \\ \\ \pi r^3 &= 150 \\ \\ r &= \sqrt[3]{\frac{150}{\pi}} \approx 3.628\text{ cm}\end{aligned}[/tex]
Hence, the radius that minimizes the surface area will be about 3.628 centimeters.
Then the height will be:
[tex]\displaystyle \begin{aligned} h&= \frac{300}{\pi\left( \sqrt[3]{\dfrac{150}{\pi}}\right)^2} \\ \\ &= \frac{60}{\pi \sqrt[3]{\dfrac{180}{\pi^2}}}\approx 7.25 6\text{ cm} \end{aligned}[/tex]
In conclusion, the dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.