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.

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) (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 , the vertex is always at .

In this case,
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

As you can see, the vertex (the highest point) of this parabola is at x= 20.

R^2 at SCC

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