# Global Extrema/Area of a Rectangle

• Quantum_Grid
In summary, to find the dimensions of the rectangle with the largest area and a perimeter of 200 meters, you need to write the area as a function of just one variable (x or y) and then find the value of that variable when the function is at its maximum. This can be done by solving for x or y in terms of the other variable using the perimeter constraint, and then using that relationship to find the value of the other variable. In this case, the dimensions of the rectangle will be 50 meters by 50 meters, creating a square with the maximum area.

## Homework Statement

Find the dimensions of of the rectangle with perimeter 200 meters that has the largest area.

## The Attempt at a Solution

This is in the section on Global Maxima/Minima so I know it has to be something with graphing a formula and finding the maxima, but I cannot figure where to start on this. The book is no help (Applied Calculus - Hughes-Hallett, et.al) and this is not the first time it has presented problems that it does not even began to describe in the text. :grumpy:

Can someone tell me where to start?

What is the equation for the perimeter of a rectangle with length x and height y (and what do you have to set this equal to)? What is the equation for the area of this rectangle?

Dunkle said:
What is the equation for the perimeter of a rectangle with length x and height y (and what do you have to set this equal to)? What is the equation for the area of this rectangle?

Ok, so I have 200=2(x+y), and the area is A=xy, right? I must be missing something terribly obvious, because I am still stuck on where to go from here, except that maybe 100=x+y. How does this fit into a Global Extrema problem?

Okay, good! Now, you need to write the area in terms of just one variable so that you can find the maximum of the area function. Any ideas?

y=A/x? Man, I can almost get this, I feel like I'm so close, but where does the perimeter function come in, y=100-x?

Is this problem an an easy one? Because it seems like it, and it's making me feel so dumb that I can't figure it out!

You are so close! You are trying to maximize the area, so you need to write the area as a function of x (or y) only. You can solve for x in terms of y (and vice versa) using the perimeter constraint. 2x+2y = 200 => y = 100-x => A = x(100-x). Now how do we find the value of x when this function is at its maximum? Once you find this x, you can solve for y using the relationship we found between x and y using the perimeter constraint. Is this clear?

50x50! Dude, thanks for your patience on that one, it was driving me crazy! It makes sense now.

You're welcome! Also, note that this answer may have been what you expected from the beginning: a square!