Another word prob.:Amanda has 400 feet of lumber to frame a

[cakers44]

Amanda has 400 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). She wants to maximize the area of her patio (area of a rectangle is length times width). What should the dimensions of the patio be, and show how the maximum area of the patio is calculated from the algebraic equation. Use the vertex form to find the maximum area.

Not sure what equation to use.... word problems are so confusing to me.

2L+ 2W = 400
L X W = area

Thanks!

 

[tkhunny]

Wow! I like that problem statementn. It provides ALL necessary information and even a means to the solution. Oh, that many, many more problems could be written so clearly.

You have only to substitute and solve.

2L+ 2W = 400 ==> 2L = 400 - 2W ==> L = 200 - W

Area(L,W) = L * W

Then, Area(W) = (200-W) * W = 200W - W^2

Can you see the quadratic form that should start you thinking about parabolas?

 

[galactus]

"wurd problem, schmurd problems". :wink:

Word problems make you think about something more like real life than an arbitrary equation to solve.

Say this 3 times, "word problems are my raison d'etre".

Anyway, you know perimeter is \(\displaystyle 2x+2y=400\)...[1]

You know area is given by \(\displaystyle A=xy\)....[2]

Solve [1] for y, sub into [2]. You will have an area equation in terms of x.

You should get a quadratic. Find the max by using the vertex thing. \(\displaystyle x=\frac{-b}{2a}\)

 

[TchrWill]

Re: Another word problem - thanks!

Amanda has 400 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). She wants to maximize the area of her patio (area of a rectangle is length times width). What should the dimensions of the patio be, and show how the maximum area of the patio is calculated from the algebraic equation. Use the vertex form to find the maximum area.

Not sure what equation to use.... word problems are so confusing to me.

2L+ 2W = 400
L X W = area

Thanks

Considering all rectangles with the same perimeter, the square encloses the greatest area.
Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.

Considering all rectangles with the same area, the square results in the smallest perimeter for a given area.