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:

Substitute:

Solve for <em>h: </em>

Recall that the surface area of a cylinder is given by:

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 <em>h</em>.

Find its derivative:

Solve for its zero(s):
![\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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20%280%29%20%26%3D%204%5Cpi%20r%20%20-%20%5Cfrac%7B600%7D%7Br%5E2%7D%20%5C%5C%20%5C%5C%204%5Cpi%20r%20-%20%5Cfrac%7B600%7D%7Br%5E2%7D%20%26%3D%200%20%5C%5C%20%5C%5C%204%5Cpi%20r%5E3%20-%20600%20%26%3D%200%20%5C%5C%20%5C%5C%20%5Cpi%20r%5E3%20%26%3D%20150%20%5C%5C%20%5C%5C%20r%20%26%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B150%7D%7B%5Cpi%7D%7D%20%5Capprox%203.628%5Ctext%7B%20cm%7D%5Cend%7Baligned%7D)
Hence, the radius that minimizes the surface area will be about 3.628 centimeters.
Then the height will be:
![\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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Cbegin%7Baligned%7D%20h%26%3D%20%5Cfrac%7B300%7D%7B%5Cpi%5Cleft%28%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B150%7D%7B%5Cpi%7D%7D%5Cright%29%5E2%7D%20%20%5C%5C%20%5C%5C%20%26%3D%20%5Cfrac%7B60%7D%7B%5Cpi%20%5Csqrt%5B3%5D%7B%5Cdfrac%7B180%7D%7B%5Cpi%5E2%7D%7D%7D%5Capprox%207.25%206%5Ctext%7B%20cm%7D%20%20%20%5Cend%7Baligned%7D)
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.
The answer is: [D]: " y = x² + 2x + 2 " .
______________________________________________
Notice the corners; the corner on each triangle with one line is equivalent, and the same goes for the corner on each triangle with two lines, and the corner on each triangle with three lines. this is just a way to know which angles match up!
this means the answer would be:
D!
Answer:
Obtuse Angle
Step-by-step explanation:
It is greater than 90˚
Rational Numbers do not include
a. infinate decimals