# Thread: In need of help.

1. ## 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

2. 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: $\displaystyle x+x<x+l$

Substitute and add: $\displaystyle 2x<16$

Finally: $\displaystyle x<8$

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.

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.

3. 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.