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
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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) (Show Source): You can put this solution on YOUR website!
Start with the perimeter formula
Plug in (the given perimeter)
Factor out the GCF 2
Divide both sides by 2
Subtract W from both sides
Rearrange the equation
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Move onto the area of a rectangle formula
Plug in
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 to get
From the graph, we can see that the max area of occurs when the width is (note: use the trace/extrema feature on your calculator).
Now simply divide the area by the width to get the length
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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.
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