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Numerade Educator



Problem 58 Hard Difficulty

Find a formula for the described function and state its domain.

A rectangle has area $ 16 m^2 $. Express the perimeter of the rectangle as a function of the length of one of its sides.


$P=\frac{2 x^{2}+32}{x}$

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Video Transcript

we want to find a function for the perimeter of the rectangle. So let's go ahead and label the rectangle with L for length and W for width. And so we know that perimeter is two times the length, plus two times the width. Now, we're supposed to find the perimeter as a function of the length of one of its sides, so we should have only one variable in here not to. So this is where the area is going to come into play. We know that area is 16 and we know that area is like times with so we can use this equation to isolate W. And we have 16 divided by l equals W so we could substitute that into our primary equation, our perimeter equation where the W is So now we have perimeter equals two l, plus two times 16 over l. That will help us to describe the perimeter as a function of only one variable. We could go ahead and simplify this, so we have to l plus 16 or plus 32 over l. And if we don't like having two separate fractions like that, I mean, I'm perfectly fine with it. But if you got your answer and you looked it up in the back the book and it looked different, you might want to know ways to simplify it. So what you could do is you could multiply the second or the first fraction by L over l to get a common denominator. So we would have to l squared over l plus 32 over l. And that gives us to l squared plus 32 over L as a way to describe the perimeter. Okay, so next we want to find the domain. So what are the possible values of L? Well, we don't know a whole lot about this rectangle. We don't know if it happens to be very, very, very narrow with a very small L and a very large W still with an area of 16. Or if it happens to have a very long, very large l in a very small w we don't know. We just know that area equals length. Times with 16 equals length, times with and that the width is 16 divided by the length. You could also say that the length is 16 divided by the width. If the width is very, very, very small. The length could be really, really huge. Imagine that the width is 0.1 The length is going to be like 16,000. So the length could just keep getting bigger and bigger and bigger if the width cut Really small. But the smallest the length could be a zero. So we do have a lower bound on that. So for the domain, we can say that the length has to be get myself a little space here somewhere. Put it up here. The domain is that the link has to be greater than zero.