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What is the area enclosed by the graph of $|3x|+|4y|=12$?

 Dec 18, 2014

Best Answer 

 #4
avatar+23254 
+10

To emphasize what Melody did:

The equation:  |3x| + |4y|  =  12

Can be split into 4 separate equations; 2 for the  |3x| = ± 3x  and  2 for the  |4y| = ± 4y :

--->                 3x + 4y  =  12       --->   y  =  (12 - 3x)/4

                      -3x + 4y  =  12       --->   y  =  (12 + 3x)/4

                       3x - 4y  =  12       --->   y  =  (12 - 3x)/-4

                      -3x - 4y  =  12        --->   y  =  (12 + 3x)/-4  

Graphing these individually will give you the parallelogram that Melody got.

 Dec 18, 2014
 #1
avatar+118703 
+10

I 'cheated' a bit because I ran off and graphed it.

Here is the graph.

 

https://www.desmos.com/calculator/rrxgm9spsh

 

I expected it to be a parallelogram and it was.

(I can talk more about this if you want me too)

The easiest way to find the area is to split it into 2 congruent trianges

base = 8 units and height = 3 units

Area =   2×(12×8×3)=24   units squared. 

 Dec 18, 2014
 #2
avatar+130466 
+5

Note the graph.......https://www.desmos.com/calculator/dbrn6oivcb

These are two triangles which have a base = 8 and a height  = 3

So, the area is just 2(1/2)(8)(3)  = 24 units

 

 Dec 18, 2014
 #3
avatar+130466 
+5

Melody and I both cheated...(but she cheated first.....and as chief mod, she should set a BETTER EXAMPLE to the people of Camelot!!!)...LOL!!!!

 

 Dec 18, 2014
 #4
avatar+23254 
+10
Best Answer

To emphasize what Melody did:

The equation:  |3x| + |4y|  =  12

Can be split into 4 separate equations; 2 for the  |3x| = ± 3x  and  2 for the  |4y| = ± 4y :

--->                 3x + 4y  =  12       --->   y  =  (12 - 3x)/4

                      -3x + 4y  =  12       --->   y  =  (12 + 3x)/4

                       3x - 4y  =  12       --->   y  =  (12 - 3x)/-4

                      -3x - 4y  =  12        --->   y  =  (12 + 3x)/-4  

Graphing these individually will give you the parallelogram that Melody got.

geno3141 Dec 18, 2014
 #5
avatar+118703 
0

Thanks Geno. :)

 Dec 18, 2014

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