The perimeter of a rectangle is fixed at 30cm30\operatorname{\mathrm{cm}}. If the length LL is increasing at a rate of 2cms2\operatorname{\frac{\mathrm{cm}}{s}}, for what value of LL does the area start to decrease? Hint: the rectangle's area starts to decrease when the rate of change for the area is less than 0.

7.5\operatorname{\mathrm{cm}}"> L > 7.5 c m ⁡ L>7.5\operatorname{\mathrm{cm}}

The perimeter of a rectangle is fixed at 30 c m 30\operatorname{\mathrm{cm}} . If the length L L is increasing at a rate of 2 c m s 2\operatorname{\frac{\mathrm{cm}}{s}} , for what value of L L does the area start to decrease? Hint: the rectangle's area starts to decrease when the rate of change for the area is less than 0.

Some of these related rates problems you're going to come across are going to be a bit tricky, and you're going to have to think a little bit outside of the box to be able to solve them. But if you can solve a more complicated related rates problem like this one, you're probably gonna be pretty solid on this process for solving related rates when we have some kind of growing or shrinking shape. So without further ado, let's get right into things. Now here we're told the perimeter of a rectangle is fixed at 30 centimeters. If the length l is increasing at a rate of 2 centimeters per second, for what value of l does the area start to decrease? So that's kinda weird, but let's go ahead and draw a picture of the situation to try and really grasp what's going on here. So we have some kind of rectangle. We know that much. And what we're told about this rectangle is that it is fixed at a perimeter of 30 centimeters. So what that means is that the all of the side lengths of this rectangle are going to always add up to be 30 centimeters. There is some kind of change of the rectangle over time, but the perimeter will always be 30 centimeters. So we have a length here and a length there. These are gonna be the same length on top and bottom, and then we have the width on the left and right side, which are also going to be the same. So this is our rectangle. Now we're told that the length l is increasing at a rate of 2 centimeters per second, so that means that the length of this rectangle is going to be increasing. And if the rate, or if the length of this rectangle is increasing, that means that the width of this rectangle is actually decreasing. And here's the reason that I know that, because the perimeter is fixed. So we couldn't have a situation where a rectangle increases, but the width stays the same. Because this perimeter would be way bigger than the perimeter that we had before. So if the length is increasing, the width has to decrease, and this has to keep happening. And what's gonna happen here is because the width is decreasing, the total area inside here is going to decrease as well. But it's not going to always be the case. There's going to be a moment that that that that area starts to decrease, and that's what we really have to find. So let's go ahead and figure this out. Now an equation that I can use to set this up is to recognize the perimeter is going to be all of the sides added together. So So that's going to be length plus length plus width plus width, which I can also write as 2 times the length plus 2 times the width is equal to our perimeter of 30. But if I tried to go ahead and go through the process of related rates by taking the time derivative and solving this all through, I would run into a roadblock here. And you could go through this process if you wanted to, but what you're gonna notice is that for this width right here, we're going to get a changing width with respect to time. The reason this is a problem is that's not a rate that's given to us. We're given the changing length, but we're not given the changing width. In fact, to go ahead and pull out all the variables that we have here, I can see that the perimeter we have is 30 centimeters. So the perimeter is 30. We have that the changing length over time, the the rate at which the length increases is 2 centimeters per second. But what we're trying to do is find when the area starts to decrease. Now this is a little bit tricky because we are looking for a d a over d t kind of, but we're not looking for a specific value. We're looking for a moment that it decreases. And the hint that we're given says the rectangle's area starts to decrease when the rate of change for the area is less than 0. So we're looking for the whole moment that this rate of change here is less than 0. So this is what we're ultimately looking for in this problem. So again, there's gonna be some steps here, but that's what we're trying to find. So because of this, I since I'm gonna get a d l over d or a d w over d t when I take the derivative here, I can't use this equation by itself. But another equation I could possibly think of is the area of a rectangle. Since we're trying to find the rate at which the area starts to decrease, well, the area of a rectangle is just gonna be length times width. But we're gonna run into another issue here because notice once again we have width in the equation and this width is gonna give us a d w over dt, which once again we do not have. So we can't just directly use the area equation here either. But what if I could combine these two equations? What if I could find a way to eliminate this w from being a problem and just be able to directly solve it since we know that the length rate is something that we actually have in this problem? Well, let's try that. So what I'm going to do is go up to this equation, and I'm going to go ahead and solve it for w. So that way we can get rid of w in this bottom equation. So I can do that by subtracting 2 l on both sides here. Now doing that, that's going to get the 2 l's to cancel on the left side of the equation. So what it's going to give me is that 2 w is equal, and I'm gonna actually move this area equation down here because we'll get to it in a moment. That's going to give me that 2 w is equal to 30 minus 2. So 2 w is equal to 30 minus 2 l, and then I can divide 2 completely on both sides of this equation. That's gonna get the twos to cancel there, leaving me with just w. And that's going to give us that w is 30 divided by 2, which is 15, and then these twos will cancel here giving us minus l. So this right here would be an expression for the width that we have. And what I can do is take this expression and swap it out for the width that we have there. So what we're going to get is that the area is equal to the length multiplied by 15 minus l. And if you think about it, well, this actually has just about everything that we need. We have the dl over dt, which is gonna come out when we take the derivative here. We know that the d a over dt is gonna come out when we take this derivative. So by using the perimeter that we were already given at the start, we were able to combine this with the area of a rectangle that we know about to get this equation that we need. So sometimes there are going to be steps in these related rates problems where we actually have to go through and kind of derive an equation. But with the information given, you will be able to do that and just think about what rates are going to pop out what variables you have in the problem in order to figure out what equation you need here. So now let's go ahead and go through this process of taking the derivative. Well, to make things a little bit more simple, I will distribute this l into the parentheses. So we're going to get that this whole thing is 15 l minus l squared. Now from here, I will take the time derivative on both sides of this equation. This is just like what we've been doing for all related rates problems that we've seen. So we'll have the derivative of area with respect to time, which is d a over d t. This whole thing is going to be equal to the derivative of 15 l. Now this l would just derive off, but using the chain rule, we're going to get d l over d t. So we're gonna get 15 d l over d t. That's going to be minus, and in this case, we're going to have 2 l using the power rule here, and that's going to be d l over d t. So this right here is going to be our derivative. Now at this point, to make things a little bit more simple, I'll factor this dl over dt rate out of this expression. So we're going to get that d a over dt is equal to dl over dt, and that's going to be times 15 minus 2 l. So that's what we get. Now let's remember again what we're trying to solve for in this problem. We're trying to solve for the situation where the rectangle's area starts to decrease, and we're asked for what at what point is this rate of change going to be to going to decrease, which is gonna be when this whole thing is less than 0. So we can see is that d a over d t here has to be less than 0, which means this whole expression has to be less than 0. So we can say that d l over d t, and then that's going to be multiplied by 15 minus 2 l. This whole thing has to be less than 0 based on what we defined right here. I'm just taking this portion, this expression for d a and I'm replacing it in right there for d a over dt. So So we have this expression, and now let's actually plug in the variables that we know we have in this problem. Well, dl over dt is given to us as 2. Then we have 15, which is going to be which is that's just the number given to us. That's going to be minus 2 times l. We don't have the length given to us yet, but that length is constantly changing. So we're looking for the instant at which this length is going to be when the area starts to decrease. Now I can divide this 2 on both sides of the equation because I know that's just going to give us 0 divided divided by 2 or 0. So we'll get 15 at minus 2 l is less than 0. And then what I can do is subtract this 15 on both sides of the equation, that's when you get the 15s to cancel there, giving me that negative 2 l is less than negative 15. Now what I can do from here is multiply both sides of the equation by negative one. But recall, whenever you multiply or divide by a negative number when solving an inequality, you have to flip the inequality symbol. So it's going to give me that 2 l is greater than 15. Now from here, I can divide this 2 on both sides of the equation, and that's going to give me that l is greater than 15 over 2. So l is greater than 15 over 2, which by the way is the same thing as 7 point 5. So this right here would be the l value that we have to be greater than in order for our sit to see the situation when the area is going to start to decrease. So a way that we could say this is that our when it asks here, if the length l is increasing at this rate, for what value of l does the area start to increase? That's going to be the value of l when it is more than 7.5 or 15 over 2 centimeters. So we're gonna have more than 7.5 or 15 over 2 centimeters, and that right there is the solution to our problem. So that is how you can solve these types of situation with related rates. And again, I'll mention, if this was a problem that you felt you could solve and follow, I think you're gonna be very solid on this concept for dealing with related rates problems. So hope you found this video helpful, and let's move on.

The perimeter of a rectangle is fixed at 30 c m ⁡ 30\operatorname{\mathrm{cm}} . If the length L L is increasing at a rate of 2 c m s ⁡ 2\operatorname{\frac{\mathrm{cm}}{s}} , for what value of L L does the area start to decrease? Hint: the rectangle's area starts to decrease when the rate of change for the area is less than 0.

7.5\operatorname{\mathrm{cm}}"> L > 7.5 c m ⁡ L>7.5\operatorname{\mathrm{cm}}

15\operatorname{\mathrm{cm}}"> L > 15 c m ⁡ L>15\operatorname{\mathrm{cm}}

0\operatorname{\mathrm{cm}}"> L > 0 c m ⁡ L>0\operatorname{\mathrm{cm}}

10\operatorname{cm}"> L > 10 cm ⁡ L>10\operatorname{cm}

4. Applications of Derivatives

Related Rates

Calculus

4. Applications of Derivatives

Related Rates

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Multiple Choice

The perimeter of a rectangle is fixed at $30 c m$ . If the length $L$ is increasing at a rate of $2 \frac{c m}{s}$ , for what value of $L$ does the area start to decrease? Hint: the rectangle's area starts to decrease when the rate of change for the area is less than 0.

$L is greater than 15$

$L is greater than 7.5$

$L is greater than 0$

$L is greater than 10 cm$

Verified step by step guidance

Start by recalling the formula for the perimeter of a rectangle, which is given by $ P = 2L + 2W $, where $ L $ is the length and $ W $ is the width. Given that the perimeter is fixed at 30 cm, we have $ 2L + 2W = 30 $.

Solve for the width $ W $ in terms of the length $ L $. From the equation $ 2L + 2W = 30 $, we can express $ W $ as $ W = 15 - L $.

The area $ A $ of the rectangle is given by $ A = L \times W $. Substitute $ W = 15 - L $ into this equation to get $ A = L(15 - L) = 15L - L^2 $.

To find when the area starts to decrease, we need to determine when the rate of change of the area $ \frac{dA}{dt} $ is less than 0. Differentiate the area function with respect to time $ t $: $ \frac{dA}{dt} = \frac{d}{dt}(15L - L^2) = 15\frac{dL}{dt} - 2L\frac{dL}{dt} $.

Given that $ \frac{dL}{dt} = 2 $ cm/s, substitute this into the derivative to get $ \frac{dA}{dt} = 15(2) - 2L(2) = 30 - 4L $. Set $ 30 - 4L < 0 $ to find when the area starts to decrease. Solve this inequality to find the critical value of $ L $.

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