Step-by-step explanation:
Can you please reword your problem
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.
Let's calculate each area.
4 m:
We need to divide it by 2 to find the radius.
4÷2=2
Now let's put it in the formula.
2²π
4π=12.5663706≈13
6 m:
We need to divide it by 2 for the radius.
6÷2=3
Now put it in the formula.
3²π
9π=28.2743≈28
Now subtract 13 from 28.
28-13=15
So, the difference between their area is 15 m².
Answer:
12566.4 mm²
Step-by-step explanation:
The Petri dish shown has no lid. What is the surface area of the outside of the Petri dish? Round to the nearest tenth. A cylindrical-shaped Petri dish with a radius of 50 millimeters and a height of 15 millimeters.
TSA = Total Surface Area
CSA = Curved Surface Area
TSA of open cylinder = CSA of cylinder + area of base or top
= 2πrh + πr²
= πr(2h + r)
From the above question:
r = 50mm
h = 15mm
Hence,
= π × 50 (2 × 15 + 50)
= π × 50 (30 + 50)
= π × 50(80)
= π × 4000
= 12566.370614mm²
Approximately = 12566.4 mm²
The Surface Area of the petri dish with no lid = 12566.4 mm²