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Alika [10]
3 years ago
7

John and Matt are going to fill a pool with 2 different sized hoses. John can fill the pool in 5 hours, while Matt can complete

it in 10 hours. Explain each step in solving this equation.
Mathematics
1 answer:
Gekata [30.6K]3 years ago
3 0
The complete question is
John and Matt are going to fill a pool with 2 different sized hoses. John can fill the pool in 5 hours, while Matt can complete it in 10 hours.How long will it take both to fill the pool? Explain each step in solving this equation.

we know that
<span>John can fill the pool in --------------> 5 hours
</span>therefore
<span>I calculate the amount of pool that John fills in one hour
</span>if John can fill 100% of the  pool in----------------> 5 hours
X--------------------------------------> 1 hour
X=1/5=0.20 pool/hour

Matt  can fill the pool in --------------> 10 hours
therefore
I calculate the amount of pool that Matt  fills in one hour
if Matt can fill 100% of the  pool in----------------> 10 hours
X--------------------------------------> 1 hour
X=1/10=0.10 pool/hour

<span>adding both amounts

(0.20+0.10)=0.30 -----------> 30% pool/hour
then
</span>if both can fills 30% of the  pool in----------------> 1 hour
100%-------------------------------> X
X=100/30=3.33 hours----------> 3 hours + 19 minutes+ 48 sec

the answer is 3.33 hours (3 hours + 19 minutes+ 48 sec)

<span>The equation to determine the amount of pool filling (y) according to time (t) in hours is given by

</span><span>y=0.30*t 

</span>
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You have a puppy named Fido and 60 feet of fencing. You want to make him the largest play area possible. What are the dimensions
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Answer:

<u>15 feet x 15 feet</u>

Step-by-step explanation:

I'm worried about Fido, so I'll use three different approaches to maximizing his play space:  Graphing, Chart, and First Derivative.  This is perhaps more than Fido wants, but it hopefully offers ideas on how these types of problems can be answered.

=========================

Fido is limited to 60 feet of fence, which will form the perimeter.  The area  of a rectangle is (length)*(width).  Let's set length to L and width to W.  I'll use A for area.

Therefore A = L*W

The rectangle's perimeter will be 2L + 2W, and this is equal to 60 feet:

2L + 2W = 60

Let's rearrange this equation to isolate either W or L.  I chose L:

L + W = 30

L = 30 - W

Now let's use this definition of L in the Area equation:

A = L*W

A = (30-W)*W

A = 30W - W^2

<u>Graphing</u>

We can graph this function (A = 30W - W^2) to find the maximum area.  See the attached image.  We find a maximum area of 225 ft^2 when the width is 15 feet.  15 feet would mean that Fido's roaming pen will be in the shape of a square;  15 feet for each side equals 60 feet total fence.

<u>Chart</u>

One can also find a maximum by calculating a variety of conditions set by L = 30 - W.  I randomly picked lengths from 2 to 24 feet and calculated the resulting width and then the resultant area for each condition.  See attached image.  Although I did not pick 15 feet, we can see that the area increases until the length transitions from 14 to 16 feet.  It reverses direction at that point. A closer inspection around those values will show that 15 feet is, in fact, the optimum value to produce a pen with the most area.

<u>First Derivative</u>

One could also take the first derivative of the equation A = 30W - W^2 and set it equal to zero.  The fist derivative tells us the slope of the line at any point (for a value of W in this example).  The slope is zero only when the curve hits a maximum.

A = 30W - W^2

A' = 30 -2W

0 = 30 - 2W

2W = 30

W = 15 feet

The slope is zero when the width is 15 feet.  15 feet is the optimum for creating a pen with the most area possible with 60 feet of fence.

===

All three approaches yield the same result.  Graphing is fun when you have access to DESMOS.  Charts are fun if you like Excel.  Derivatives can also be fun, mostly because I always ponder "Who had the audacity to think this might work?"  (Didn't they have something more important to do?  Raid a village, etc.?).

====

Arf

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