SOLUTION: Area and parameter of rectangle is 6000 and 340 then length of parameters

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Question 1129657: Area and parameter of rectangle is 6000 and 340 then length of parameters
Found 2 solutions by solver91311, MathLover1:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


You used the word "parameter" twice, once completely incorrectly and once in a very confusing way. I presume you meant to say: The area of a rectangle is 6000 square units and the perimeter is 340 units. What are the measures of the length and width? If English is a second language for you, please try to get help from someone more fluent before posting the next time.

Let and represent the dimensions of the rectangle. Then:

Area = and Perimeter = . Hence:





So





Put the quadratic into standard form and solve for , then calculate


John

My calculator said it, I believe it, that settles it

Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!

Area of rectangle is 6000:

L%2AW=6000

W=6000%2FL........eq.1



and perimeter is 340+:
2%28L%2BW%29=340
L%2BW=340%2F2
L%2BW=170
W=170-L.......eq.2


from eq.1 and eq.2

6000%2FL=170-L
6000=170L-L%5E2
L%5E2-170L%2B6000=0
L%5E2-120L-50L%2B6000=0
%28L%5E2-120L%29-%2850L-6000%29=0
L%28L-120%29-50%28L-120%29=0
%28L+-+120%29+%28L+-+50%29+=+0
solutions:
if %28L+-+120%29+=+0=>L=120
if +%28L+-+50%29+=+0=>L=50

so, L=120 or L=50

then
W=170-L=>W=170-120=>W=50
W=170-L=>W=170-50=>W=120
=>W=50 or W=120

since length is greater than width, then length of rectangle is L=120 and the width is W=50