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Stells [14]
3 years ago
5

Question part points submissions used a fence 8 ft tall runs parallel to a tall building at a distance of 4 ft from the building

. what is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building? (round your answer to two decimal places.)
Mathematics
1 answer:
denpristay [2]3 years ago
3 0
This is a good question. also one you can use in real life.

if the fence is 8ft tall you know you need at least an 8ft ladder to get over the fence.

the building is 4ft away but you need to clear the fence.

|\
|™\
| ™\
|__\
|™™ | \
|.....|...\........

So this is what your looking for. I find it easier to learn when you make it into a picture.

so it's 8ft high and 4ft from building so the ladder has to be 8×4 to go from the ground to the wall over the fence.
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A rectangular storage container with an open top is to have a volume of 24 cubic meters. The length of its base is twice the wid
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Answer:

419.25

Step-by-step explanation:

The calculation of the cost of materials for the cheapest such container is shown below:-

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Height = h

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= 2x^2

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= 4xh + 2xh

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h = \frac{24}{2x^2} \\\\ h = \frac{12}{x^2}

Now, cost is

= 13(2x^2) + 9(4xh + 2xh)\\\\ = 13(2x^2) + 9(4x + 2x)\times \frac{12}{x^2} \\\\ = 26x^2 + \frac{648}{x}

now we have to minimize C(x)

So, we need to compute the C'(x)

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C"(x)  = 52x - \frac{1,296}{x^3}

now for the critical points, we will solve the equation C'(x) = 0

= 52x - \frac{648}{x^2} = 0\\\\ x = \frac{648}{52}^{\frac{1}{3}}

C" = ((\frac{648}{52} ^{\frac{1}{3} } = 52 + \frac{1296}{(\frac{648}{52})^\frac{1}{3} )^3}\\\\ = 52 + \frac{1296}{\frac{648}{52} } >0

So, x is a point of minima that is

= (\frac{648}{52} )^\frac{1}{3}

Now, Base material cost is

= 13(2x^2)\\\\ = 26(\frac{648}{52} )^\frac{2}{3}

= 139.75

Side material cost is

= \frac{648}{x} \\\\ = \frac{648}{(\frac{648}{52})^\frac{1}{3}  }

= 279.50

and finally

Total cost is

= 139.75 + 279.50

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