# Differentiation word problem (basic)

1. Jan 9, 2013

### Taylor_1989

A rectangle of length x, where x varies, has a constant area of 48cm2. Express the perimeter, y in terms of x. Find the least possible value of x.

my problem is not the maths part i.e. the differentiation, but the equation to get things moving. I really have no idea where to start. I drew a rectangle, but am not sure if I am suppose to call the width by a letter. What's really confusing me is the "x varies", what dose this mean? All I see I have an x and a 48 which to me, I cant seem to connect with the perimeter.

I would really appreciate some guidance here if possible. Thanks in advance.

2. Jan 9, 2013

### Saitama

Assume the width of rectangle as b.
Area, xb=48.
Perimeter, y=2(x+b).
Substitute b in the second equation.

3. Jan 9, 2013

### CAF123

If the width of the rectangle multiplied by it's length is always a constant, as the length varies (say increases) then it's width must decrease so as to preserve this constant area.

4. Jan 9, 2013

### Taylor_1989

Thanks for the reply final, got my ans in the end $16\sqrt3$ definitely going to need more practice with these. Has anyone got any tips, or is something that if you do enough of them you get used to them?

5. Jan 9, 2013

### Staff: Mentor

Well, yes, this is usually the difficult part, and this is part of mathematics, being able to analyze a situation and come up with an equation that represents it. After you have an equation, then you don't really have to think as much.
As you do more of them, you start to develop organization and analysis skills, and they tend to get a little easier.

For problems like this, you started off well by drawing a picture, but didn't follow through. You have a rectangle whose length is x, so hopefully you drew a rectangle and labelled its length as x. You are given that the area is 48.

Since you're not given the width, label the width of the rectangle - I would use w instead of b, but that's not very important. I definitely wouldn't use y, because the problem statement talks about y as the perimeter. (A better choice would be P, though.)

After you have the rectangle labelled, write equations for the perimeter and area.

You can extend these ideas to other problems - look carefully at the problem statement and translate the words into equations.

6. Jan 10, 2013

### Ray Vickson

Your later post indicates you have found the answer, but I hope you notice that your answered the *correct* question, rather than the one asked above. There is no least possible value of x; there is, though, a least possible value of perimeter (I hate using the letter y for that).