A rancher has 48 meters of wire to make a pen for his cows...

eddy2017

eddy2017

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Hi, by now I know you know i hate to be in doubt and not be able to find an explanation for a mathematical reasoning applied to something. This might a stupid problem, but i am not getting something here and i will ask you.

A rancher has 48 meters of wire to make a pen for his cows. Since his property is next to a river, what is the largest rectangular area he can fence if he uses the river as one side for the enclosure?.
this is the answer given:

In order to solve this type of problems we apply a division.

48 mts total

15 * 2 = 30 mts

9 * 2 = 18 mts

This would be the largest area you can fence with that length to make a rectangle 15 mts and 9 mts.
I do not get the reasoning.
Only that 15+9=48 but why the multiplication by 2 of each number?.
Thanks,
eddy
 
I disagree with the “given answer”

if the rancher uses the river as one side …

[imath]2W+L = 48[/imath]

[imath]A = LW = (48-2W)W = 48W-2W^2[/imath]

the max area would occur when [imath]W = \dfrac{-48}{-4} = 12[/imath]
 
skeeter said:
I disagree with the “given answer”

if the rancher uses the river as one side …

[imath]2W+L = 48[/imath]

[imath]A = LW = (48-2W)W = 48W-2W^2[/imath]

the max area would occur when [imath]W = \dfrac{-48}{-4} = 12[/imath]
Click to expand...
Thanks, skeeter.
 
Subhotosh Khan said:
How is that?
Click to expand...
No, it was the wrong thing to say. What I wanted to say " not only that 15+9 =48 but also that....
Don't get why 15+ 9 either.
Other than the idea of finding two numbers that when multiplied by 2 their individual results added to 48.
 
Last edited:
eddy2017 said:
No, it was the wrong thing to say. What I wanted to say " not only that 15+9 =48 but also that....
Don't get why 15+ 9 either.
Other than the idea of finding two numbers that when multiplied by 2 their individual results added to 48.
Click to expand...
Before you hit the "Post reply" button - please review your response - possibly twice to eliminate Khan-fusion!!!!
 
A Hispanic teacher explained another way to say this, but for the life of me i did not make head or tail of it. i am bringing it here just as he explained it to see if you can help me decipher all this and understand the why of some things he explained. This is a totally different way that skeeter used, which for me it was way more clear.
my rough translation of what he said and did:
Well, here it is:

The area measure S of the rectangle is xy. S is a function of two variables. S(x;y)=xy
We know the equation 2x+y=48
We can therefore calculate y as a function of x.
y=48-2x
so S is a function of the single variable x.
Once S(x) is calculated, we look for its maximum. Either by calculating the derivative S' and drawing up the table of variations of S, or, S being a second degree trinomial of the variable x, by returning to its canonical form.

ile_TEX.cgi

ile_TEX.cgi

ile_TEX.cgi


The maximum area is 288 SINCE x=12 and y=24

I don't understand what he did.
 
looks like he completed the square with [imath]x^2 - 24x[/imath]

[imath]-2(x^2 - 24x) = -2(x^2 - 24x + 144 - 144) = -2[(x-12)^2 - 144] = -2(x-12)^2 + 288[/imath]

note the vertex form of a quadratic is [imath]y = a(x - h)^2 + k[/imath], where [imath](h,k)[/imath] are the coordinates of the vertex (a maximum in this case) ...

[imath]h=12[/imath] is the value that maximizes the area, and [imath]k=288[/imath] is the value of the maximum area.



When I wrote the equation [math]A = 48W-2W^2[/math]I just found the value of [imath]W[/imath] for the vertex ... [math]W = \dfrac{-b}{2a} = \dfrac{-48}{-4} = 12[/math]
 
Okay, thanks a lot skeeter. It is clearer now with your explanation.
 
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