1. ## Pre-Calc

A rectangle has a perimeter of 80 cm. If its width is x, express its length and its area in terms of x. What is the maximum area of the rectangle?

2. Originally Posted by ep78
A rectangle has a perimeter of 80 cm. If its width is x, express its length and its area in terms of x. What is the maximum area of the rectangle?
Let x be the width and y be the height. Thus 2x+2y=80. Solving for y, we get x+y=40 and y=40-x.

Let A(x) be an area of function. A(x) = x(40-x). That's just the length times the width all in terms of x.

Now differentiate and find the critical point.

3. Originally Posted by Jameson
Let x be the width and y be the height. Thus 2x+2y=80. Solving for y, we get x+y=40 and y=40-x.

Let A(x) be an area of function. A(x) = x(40-x). That's just the length times the width all in terms of x.

Now differentiate and find the critical point.

so would i plug in any number into x(40-x) or is there a certain one?

4. Ooops! This is pre-calc. I was throwing in some Calculus terms.

You could graph this area function and visually spot the maximum. I will say that a square is the best shape at maximizing area when the perimeter is constrained.

5. oh haha, okay thank you very much!

6. Originally Posted by ep78
so would i plug in any number into x(40-x) or is there a certain one?
Differentiate then find the critical numbers. (Where the derivative = 0)

dy/dx=40-2x

40-2x=0

x=20

7. Originally Posted by Len
Differentiate then find the critical numbers. (Where the derivative = 0)

dy/dx=40-2x

40-2x=0

x=20
Look at title. I made the same mistake. (Pre-calc)

8. Originally Posted by Jameson
Look at title. I made the same mistake. (Pre-calc)
Ahh sorry my internet crashed when I was replying and when it came back online I pressed refresh and it sent.