SOLUTION: The perimeter of a triangle cannot be more than 57 feet. If one side of the triangle is twice as long as a second side and the third side is 21 feet​ long, find the longest pos

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Question 1135084: The perimeter of a triangle cannot be more than
57 feet. If one side of the triangle is twice as long as a second side and the third side is 21 feet​ long, find the longest possible values for the other two sides of the triangle.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39614) About Me  (Show Source):
You can put this solution on YOUR website!
Two sides of x and 2x, and the last side of 21 feet.
x%2B2x%2B21%3C=57
which should be clear and not too tough to solve.

Answer by ikleyn(52756) About Me  (Show Source):
You can put this solution on YOUR website!
.
As other tutor instructed you, you have this inequality

    x + 2x + 21 <= 57      (1)

for the "second" side length x.



From this inequality, you immediately get

    3x <= 57 - 21 = 36,   so  x <= 36/3 = 12.

It answers the problem's question, but it is not a COMPLETE DESCRIPTION of possible triangles.



For example,  the sides  1, 2 and 21  satisfy the basic given inequality (1),

but they  DO NOT FORM any triangle - since they do not satisfy another triangle inequality 

    x + 2x > 21.      (2)


The inequality (2) requires  3x > 21,  which implies  x > 7.


So, only if the compound inequality holds

    7 < x <= 12,

the described triangle does exist.

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The lesson to learn from my post is THIS : 

    When you solve problems for triangle side lengths,  you must satisfy and check ALL TRIANGLE inequalities.