SOLUTION: Id need help here please, find the dimensions of a rectangle "a" with the greatest area whose perimeter is 30feet. I know that the perimeter of a rectangle is 2l+2w=P and i also kn

Algebra ->  Rectangles -> SOLUTION: Id need help here please, find the dimensions of a rectangle "a" with the greatest area whose perimeter is 30feet. I know that the perimeter of a rectangle is 2l+2w=P and i also kn      Log On


   

Question 204292: Id need help here please, find the dimensions of a rectangle "a" with the greatest area whose perimeter is 30feet. I know that the perimeter of a rectangle is 2l+2w=P and i also know that to solve for the length; l= p-2w/2 and to solve for the Area; A=lw but here i dont have the length or width so how would I solve this problem?? I have came as far as; l=p-2w/2 so l=30(ft)-2w/2... I now i didnt come far and that its pretty lousy -_- that i only came as far as that, but thats exactly why i need help.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
P=2L%2B2W Start with the perimeter formula


30=2L%2B2W Plug in P=30 (the given perimeter)


30=2%28L%2BW%29 Factor out the GCF 2


15=L%2BW Divide both sides by 2


15-W=L Subtract W from both sides


L=15-W Rearrange the equation


---------------------------


A=LW Move onto the area of a rectangle formula


A=%2815-W%29W Plug in L=15-W


A=W%2815-W%29 Rearrange the terms.


Now the goal is to maximize "A". You can think of "A" as a function of "W". In other words, the area "A" is dependent on "W". There are a number of ways to maximize "A", but the easiest way is to graph A=W%2815-W%29 to get


+graph%28+500%2C+500%2C++-2%2C+20%2C+-5%2C+60%2C+x%2A%2815-x%29+%29+


From the graph, we can see that the max area of A=56.25 occurs when the width is W=7.5 (note: use the trace/extrema feature on your calculator).


Now simply divide the area by the width to get the length L=A%2FW=56.25%2F7.5=7.5

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Answer:


So the max area or 56.25 square feet occurs when the dimensions are 7.5 feet by 7.5 feet.


Note: this is important to notice that the max area occurs when the figure is a square. So for any given fixed perimeter of a rectangle, the max area will result from a square. This may come in handy when you're trying to get the most out of your yard (in terms of area) and keep the cost of fencing to a minimum.