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LUCKY_DIMON [66]
3 years ago
14

Help if possible.

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
4 0
A ) Length = 5 - 2 x
Width = 3 - 2 x
The area of the bottom:
A = L  x  W = ( 5 - 2 x ) ( 3 - 2 x ) = 15 - 10 x - 6 x + 4 x² = 
= 4 x² - 16 x + 15
b ) 4 x² - 16 x + 15 = 10
4 x² - 16 x + 5 = 0
x = (16 - √(256 - 80))/ 8 = ( 16 - 13.266 ) / 2
x = 0.34 in
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Three rectangular tables are placed end to end to form one long table 7.8 meters long. The lengths of two of the tables are 3.5
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g A manufacturer is making cylindrical cans that hold 300 cm3. The dimensions of the can are not mandated, so to save manufactur
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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:

\displaystyle V = \pi r^2h

Substitute:

\displaystyle (300) = \pi r^2 h

Solve for <em>h: </em>

\displaystyle \frac{300}{\pi r^2} = h

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

\displaystyle A = 2\pi r^2 + 2\pi rh

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>.

\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}

Find its derivative:

\displaystyle A' = 4\pi r - \frac{600}{r^2}

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}

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}

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.

7 0
3 years ago
Need help on both of these please I am really stressed out
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Answer:

sheeeeeeeeeeshshshshshdh

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