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meriva
3 years ago
6

Math question down below

Mathematics
1 answer:
Bas_tet [7]3 years ago
4 0
I would say 25 is the answer to it
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All boxes with a square​ base, an open​ top, and a volume of 60 ft cubed have a surface area given by ​S(x)equalsx squared plus
Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval (0,\infty) is S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2

The dimensions of the box with minimum surface​ area are: the base edge x=2\sqrt[3]{15}\:ft and the height h=\sqrt[3]{15} \:ft

Step-by-step explanation:

We are given the surface area of a box S(x)=x^2+\frac{240}{x} where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval (0,\infty) and the dimensions of the box with minimum surface​ area.

1. To find the absolute minimum you must find the derivative of the surface area (S'(x)) and find the critical points of the derivative (S'(x)=0).

\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}

Next,

2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120

There is a undefined solution x=0 and a real solution x=2\sqrt[3]{15}. These point divide the number line into two intervals (0,2\sqrt[3]{15}) and (2\sqrt[3]{15}, \infty)

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before x=2\sqrt[3]{15}, increases after it, and is defined at x=2\sqrt[3]{15}. So f(x) has a relative minimum point at x=2\sqrt[3]{15}.

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for x=2\sqrt[3]{15}.

\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}

and for x=2\sqrt[3]{15} we get:

2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0

Therefore S(x) has a minimum at x=2\sqrt[3]{15} which is:

S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2

2. To find the third dimension of the box with minimum surface​ area:

We know that the volume is 60 ft^3 and the volume of a box with a square base is V=x^2h, we solve for h

h=\frac{V}{x^2}

Substituting V = 60 ft^3 and x=2\sqrt[3]{15}

h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft

The dimension are the base edge x=2\sqrt[3]{15}\:ft and the height h=\sqrt[3]{15} \:ft

6 0
3 years ago
A tape diagram. There are 65 students out of 100 students who ride the bus. There are question mark students out of 800 students
SVEN [57.7K]

Answer:

<em>800÷100 = 8</em>

<em>65×8=520</em>

\frac{520}{800}

another way:

<em>100%-65%=35%</em>

<em>35% = 0.35</em>

<em>800×0.35= 280</em>

<em>800-280=</em>520

another way:

<em>65% = 0.65</em>

<em>0.65×800=</em>520

Hope it helped! :)

6 0
2 years ago
14. An engineer is designing a circular base for a
Flauer [41]

Answer:

c=2(3.14)r

r= 5inches

c=2•3.14•5

c=31.4 inches

8 0
3 years ago
Which system of equations would have no
jenyasd209 [6]

Answer:

hi it's

Easy B

Step-by-step explanation:

No Problem

8 0
2 years ago
Which must be true select three options
tatiyna
Wait what
I don’t see anything
8 0
3 years ago
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