Question 73454: Can you help with this problem?
Find the dimensions of a rectangle "a" with the greatest area whose perimeter is 30 feet.
thank you
Answer by funmath(2933)   (Show Source):
You can put this solution on YOUR website! Find the dimensions of a rectangle "a" with the greatest area whose perimeter is 30 feet.
Let width (W) be: x
Let length (L) be:y
Perimeter (P) is:30 ft.
The formula for the perimeter is 
 Solve for either x or y, I'm going with y.
Since length (L)=y, then length now = 15-x.
Width (W) is still: x
The formula for the area is: 
If you're taking calculus you would take the derivative of this formula. Let me know if you need to solve it that way.
There are a couple of ways to solve this using algebra, if you're supposed to complete the square, let me know.
The area will be greatest at the vertex of this parabola. We can find the x value of the vertex using this formula: 
First we need to put the equation in this form:  , so that we can identify a and b.
 a=-1 and b=15, (c=0, but we don't need it for this.)
x=7.5 ft
The maximum area would be:
A=56.25 ft^2
The dimensions would be:
width: x=7.5 ft
Length: y=15-x=15-7.5=7.5 ft
These almost always work out to be squares, unless one of the sides is against a river or a house. With that in mind, just for fun, think of an easier way to solve this. (I didn't show you that way because it doesn't make sense to everyone. It has to do with finding the half-way point between the zeros.)
Happy Cacluating!!!
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