SOLUTION: Solve this problem The length and width of a rectangle must have a sum of 40 feet. Find the dimensions of the rectangle whose area is as large as possible.

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Question 21789: Solve this problem
The length and width of a rectangle must have a sum of 40 feet. Find the dimensions of the rectangle whose area is as large as possible.
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = width of the rectangle
40-x = length of the rectangle

Area = W*L
A = x(40-x)
A = 40x - x^2
A = -x^2 + 40x

This graph is a parabola (because of the x^2) that opens downward (because of teh negative coefficient of x^2), and you need to find the HIGHEST POINT, which is the vertex of the graph.

For a parabola y+=+ax%5E2+%2B+bx+%2B+c, the vertex is always at x=+-b%2F%282a%29.

In this case, x=+-b%2F%282a%29+=+-40%2F%28-2%29+=+20
width = x = 20 feet
length = 40-x = 20 feet

It makes sense because a rectangle of maximum area is a square.

It can also be solved by graphing methods. Find the highest point on the graph y+=+-x%5E2+%2B+40x
graph+%28300%2C300%2C+-50%2C+50%2C+-100%2C+500%2C+-x%5E2+%2B+40x%29+
As you can see, the vertex (the highest point) of this parabola is at x= 20.

R^2 at SCC