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Numerade Educator



Problem 4 Easy Difficulty

The length of a rectangle is increasing at a rate of $ 8 cm/s $ and its width is increasing at a rate of $ 3 cm/s. $ When the length is $ 20 cm $ and the width is $ 10 cm, $ how fast is the area of the rectangle increasing?


140 $\mathrm{cm}^{2} / \mathrm{s}$

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Video Transcript

And this question we're asked that, given the rate of change of the length and the width of a rectangle, we want to find the length and the width of the, We want to find the rate of change of the area at a certain length and a certain width. Let's remember the formula for the area of a rectangle. We know that the area is equal to the length times the width, but All three of these are functions of time, so it's more correct to say that as a function of time, the area is, the length is a function of time, multiplied by the width times a function of time. Okay people, now that we have this information, we can go ahead and straight up, apply implicit differentiation and differentiate both sides with respect to time. Now the derivative of the area with respect to time is the rate of change of the area. And we know that that's going to be equal to the right hand side, which we're going to have to use the product rule which states to keep the first function and take the derivative of the second function and add it with the second function times the derivative of the first function. Now we know that were given to fixed lengths and width. That's these two numbers here. So let's plug these in. We know that the length is 20 And the width is 10. Yeah. And we're going to multiply that by the rate of change of the length and the rate of change of the width, Which is going to be three times eight. And when we add and when we simplify this together, we're going to get 60 plus 80 which is equal to 1 40 centimetres squared her second. And that's the answer to this question.