1. ## PreCal Rectangle

Alright guys... So I was given a problem on a test for extra credit, but couldn't solve it. Regardless, I won't let it go. So here it is:
You are given a narrow piece of wood that is 90cm long. You want to split the 90cm of wood into 4 pieces so as to make a rectangular frame.
Let x be the length of a short side of the rectangle. Then x is also the length of a short piece of wood.
Write an expression for the length of a long side, then a function of x that gives the rectangles area.

Now I got this: l= 45-x, but I'm pretty sure it is wrong because it doesn't fit into the function for the area. I think a formula for maximum volume would help, but I can't remember it.

Anyway, I would really appreciate it if someone could show me how to find the long side. Thanks!

2. Originally Posted by zacharyrod
Alright guys... So I was given a problem on a test for extra credit, but couldn't solve it. Regardless, I won't let it go. So here it is:
You are given a narrow piece of wood that is 90cm long. You want to split the 90cm of wood into 4 pieces so as to make a rectangular frame.
Let x be the length of a short side of the rectangle. Then x is also the length of a short piece of wood.
Write an expression for the length of a long side, then a function of x that gives the rectangles area.

Now I got this: l= 45-x, but I'm pretty sure it is wrong <<<<<<< that's wrong: Your equation is OK
because it doesn't fit into the function for the area. I think a formula for maximum volume would help, but I can't remember it.

Anyway, I would really appreciate it if someone could show me how to find the long side. Thanks!
The area of a rectangle is calculated by:

$A = length \cdot width$

You got: w = x and l = 45 - x. Therefore

$A(x)=x(45-x)~\implies~A(x)=45x-x^2$

That means:
• A is a parabola opening down with the maximum at it's vertex

• The maximum occurs if A'(x) = 0

I don't know which method you prefer to answer the question.

3. Originally Posted by earboth
The area of a rectangle is calculated by:

$A = length \cdot width$

You got: w = x and l = 45 - x. Therefore

$A(x)=x(45-x)~\implies~A(x)=45x-x^2$

That means:
• A is a parabola opening down with the maximum at it's vertex

• The maximum occurs if A'(x) = 0

I don't know which method you prefer to answer the question.
Thanks for helping me realize that!