In need of help.

• Jan 27th 2007, 01:38 PM
Dewitt
In need of help.
Hello everyone, im new to the forums and have finally decided to join a math forum since I feel it will help me better understand math and also give me help with certain problems I have trouble with. So I wanted to see if I could get a little help with this equation since im bad with numbers :)

Equation: A string 32 inches long has been cut into 4 pieces to form a rectangle whose shortest side is X.

1.) Express the area A(x), of the rectangle as a function of X.

2.) Determine the domain of A(x).

3.) Approximate the maximun area of the rectangle and state the dimensions of the rectangle.

This is the problem I am having alot of trouble with as I have no Idea where to even start. Any help anyone could bring to this equation for me would be greatly appreciated, and thnx :)
• Jan 27th 2007, 01:51 PM
Quick
Quote:

Originally Posted by Dewitt
Hello everyone, im new to the forums and have finally decided to join a math forum since I feel it will help me better understand math and also give me help with certain problems I have trouble with. So I wanted to see if I could get a little help with this equation since im bad with numbers :)

Equation: A string 32 inches long has been cut into 4 pieces to form a rectangle whose shortest side is X.

We'll call the longer side \$\displaystyle l\$

So: \$\displaystyle 2x+2l=32\$

Then: \$\displaystyle 2(x+l)=32\$

Therefore: \$\displaystyle x+l=16\$

We know that \$\displaystyle x\$ is less than \$\displaystyle l\$, which means that: \$\displaystyle x+x<x+l\$

Finally: \$\displaystyle x<8\$

Quote:

1.) Express the area A(x), of the rectangle as a function of X.
You know that: \$\displaystyle A=xl\$

Substitute: \$\displaystyle A=x(16-x)\$

And there it is.

Quote:

2.) Determine the domain of A(x).
3.) Approximate the maximun area of the rectangle and state the dimensions of the rectangle.
Well \$\displaystyle x\$ has to be greater than 0, and that \$\displaystyle x\$ is less than 8.

Therefore: \$\displaystyle 0(16-0)<A<8(16-8)\$

Subtract: \$\displaystyle 0(16)<A<8(8)\$

Multiply: \$\displaystyle 0<A<64\$

So A(x) is between zero and sixty-four.
• Jan 27th 2007, 05:52 PM
ticbol
Quote:

Originally Posted by Dewitt
Hello everyone, im new to the forums and have finally decided to join a math forum since I feel it will help me better understand math and also give me help with certain problems I have trouble with. So I wanted to see if I could get a little help with this equation since im bad with numbers :)

Equation: A string 32 inches long has been cut into 4 pieces to form a rectangle whose shortest side is X.

1.) Express the area A(x), of the rectangle as a function of X.

2.) Determine the domain of A(x).

3.) Approximate the maximun area of the rectangle and state the dimensions of the rectangle.

This is the problem I am having alot of trouble with as I have no Idea where to even start. Any help anyone could bring to this equation for me would be greatly appreciated, and thnx :)

Here is one way.

Pre-Calculus. No differentiation/derivatives yet? Then we use parabolas for maximum A(x).

A rectangle is a quadrilateral of whose opposite sides are equal in lengths and whose 4 interior angles are all right angles.

1.) Express the area A(x), of the rectangle as a function of X.

Since the shorter side is x, and the perimeter is 32 inches, then the longer side is
(32 -2x)/2 = (16-x) in.

A(x) = x(16 -x) = 16x -x^2 ------------answer.

---------------------------------------------------
2.) Determine the domain of A(x).

A(x) = x(16 -x)
The domain of A(x) is from x=0 up to x=16, or [0,16]. -----------answer.

------------------------------------------------------
3.) Approximate the maximun area of the rectangle and state the dimensions of the rectangle.

A(x) = 16x -x^2 is a vertical parabola that opens downward---because the coefficient of the x^2 is negative.
The maximum value is the highest point or vertex of the parabola.

Let us examine the graph of A(x) = 16x -x^2
A vertical parabola whose vertex is at (h,k), that opens downward, is
(y-k) = -a(x-h)
So,
Let y = A(x)
y = 16x -x^2
y = -(x^2 -16x)
y = -(x^2 -16x +64 -64)
y = -(x^2 -16x +64) -(-64)
y -64 = -1(x -8)^2 ------------------the parabola of A(x).
Hence, point (8,64) is the vertex.
Meaning, when x=8 in., the area of the rectangle is maximum and it is 64 sq.in.

Therefore, maximum area of the rectangle is 64 sq.in. ---------answer.
when x=8 in.

The dimensions of the rectangle when maximum:
x = 8 in.
(16 -x) = (16 -8) = 8 in. also
Therefore, when area is maximum, the rectangle is a square of 8 inches per side. -----------anwser.