Turning a corner with a pole
What is the length of the l...

Question

Turning a corner with a pole

What is the length of the longest pole that can be carried horizontally around a corner at which a corridor that is a ft wide and a corridor that is b ft wide meet at right angles?

Accepted Answer

Welcome back, everyone. In this problem, a furniture mover needs to navigate a sofa around a right-angled turn where one corridor is 6 ft wide and the other is 8 ft wide. What is the maximum length of the sofa that can be moved around the corner without tilting it? Run the answer to 2 decimal places. For our answer choices, A says that it's 16.47 ft, B, 18.22 ft, C, 19.73 ft, and D 20.85 ft. Now let's try to visualize our problem here. OK, so let's draw our corner. OK, it's probably looks something like this. And for our corner, OK, let's draw our sofa. And now we know that the length of the longest sofa that can be carried around the corner is the shortest length of the line segment that connects the outer walls of the corridor. And at the same time, this segment should touch this corner as shown, OK? Now, we also know, like we said, that one corridor is 6 ft wide. OK. So let's put our 6 6 ft wide corridor down here, and then the other is 8 ft wide. Let's put it up here. OK. And no, from the end of the corridor to the end of the sofa, at the bottom of the diagram, let's call that X. And at the top of the diagram, let's call that Y, OK? Now, we know then that we're going to have two lengths in our problem. We're gonna have L1 from the first edge of, from the first end of the sofa to the segment, and then L2 from the second end of the sofa to the segment, OK? Where we know or we can say that the total length. Is And L equals L1 plus L2. Now what are we trying to really do in our problem? Well, what we, what we want to do basically, like we said, we'd like to minimize, OK, our value of L. We'd like to minimize the length of the longest sofa that can be carried through the corner to be the shortest length of the line segment. That's why we're minimizing and thus in that case, before we can minimize L, we'll need to express L in terms of one of the variables, either X or Y. Or could we do that? Well, first, let's try to express L1 and L2 in terms of L and Y. Notice we have two right triangles formed here so we can apply the Pythagorean theorem. And by that theorem. OK, by the Pythagorean theorem. Then we know that L1 is gonna be equal to the square root of X2 + 82, while L2. Is going to be equal to the square root. Make some space here. The square root Of Y2 + 62. Now we have two variables X and Y, so can we express one in terms of the other? Well, again, if we go back to our problem, we also noticed that the triangle with hypertens L2 is similar to the triangle with hypertens L1. So if we use the idea of sim similar triangles here. By similar triangles, then we know the ratios of the sides will be the same, which means that the ratio of Y 6 is going to be equal to the ratio of 8 to X. And in that case that means we can express y in terms of X. Y is going to be equal to 6 multiplied by 8 divided by X. I know we're going to get, we could multiply 6 by 8 here, but we're going to leave it like this on purpose for no to help us simplify a little later on. But now that we have this expression for Y in terms of X, that means. We could rewrite L, OK. Purely in terms of XL as a function of X, and that is gonna make it be equal to the square root of X2 plus 82. We know that that is, well, let me write 82 for no plus the square root of 6 multiplied by 8 divided by X. Squared plus 6 squared. Now, let's go ahead and try to simplify. Our first term is going to be X2 + 64, and in our second term we could rewrite our first expression as 62 divided by x2 multiplied by 82. Essentially that's what it would be plus 6 squared multiplied by x2 divided by X2. Now why would we do that? Well, notice it when we write it this way we can factor out 6 square divided by X2. OK, in our 2nd term. To make our second term be equal to the square root of 62 divided by x2 or 6 divided by X multiplied by the square root of X2 + 64, and now we can factor out root X2 + 64 to give us L being equal to the square root of X2 + 64 multiplied by 1 + 6 divided by X. Know that we have L in terms of X, we can minimize L, OK, for X greater than 0. OK. So let's go ahead and do that. How do we minimize? Well, remember that to minimize, we can find the first derivative of our function, set it to zero, solve for the critical point, and then test whether that critical point is a maximum or minimum. So let's find the derivative of L with respect to X. Now if you notice here, L is a product, so we can apply the product rule of differentiation, and in that case, we can keep our first term, X2 + 64, sorry, the square root of X + 64, differentiate the second term, which would be negative 6 divided by X2. Plus The second term, 1 + 6 divided by X, multiplied by the derivative of the first term, which would be X divided by the square root of X squared + 64. And know all of this to minimize it or to find a critical point we set all of this equal to 0. In that case, notice here that we can bring one term to the other side, bring our first term to the other side, to then say. That 1 + 6 divided by X. Multiplied by X divided by the square root of X2 + 64. It is going to be equal to 6 multiplied by root X2 + 64 divided by X2. If we multiply through by the square root of X2 + 64 and X2d, then we should get 1 + 6 divided by X. Multiplied by X cubed to be equal to 6 multiplied by x2 + 64. Expanding our terms, we'll get X cubed plus 6X2 to be equal to 6 X2 + 384, subtracting 6X2d from both sides. Then we should get X cubed to be equal to 384, which implies then that X is going to be equal to the cube root of 384, which would be approximately 7.27. So now we know that this value is a critical point for our function. Now we just need to establish if it's going to be a minimum or a maximum, and we can test that by figuring out the sign of the first derivative. Now, for the first derivative, if the sign changes from negative to positive as X increases to the critical point, then the critical point is a local minimum. So let's test that, OK? Let's test. Weather X. Is a minimum using the first derivative test. Now we know that X is approximately 7.27. So for the value of X that's less than the cube root of 384 at a critical point. If we let X be equal to 7. Then the first derivative, OK, is going to be equal to. Remember we had our expression for our first derivative here, so we just need to substitute. Oops sorry, we just need to substitute our value of 7 here. So that's minus 6 multiplied by the square root of 72 + 64 divided by 72 + 1 + 67, OK. Multiplied by 7 divided by the square root of 72 + 64. I know when we evaluate that, then we should get the first derivative when X is 7 to be approximately negative 0.079. No, for X greater than the critical point. OK. Then we can choose X to be equal to 8, and then we do that, I'll leave the working to you, but essentially you just plug it into our formula again, and we should get the first derivative at that point to be approximately 0.177. Since our sand goes from negative. To positive Then this tells us that X is indeed a minimum point or X corresponds to the minimum point. X corresponds to the minimum. I know that we have this value for X, we can go ahead and plug it in to find our, our length. Remember, the length is what we want to solve for the longest sofa that can move around the corner. So now, plugging it into our formula for L. So actually, let me write this in red. Then L is going to be equal to. The square root of X2, so that's the root of 384 squared plus 64. Multiplied by 1 + 6 divided by the cube root of 384. And that is gonna be approximately equal to 19.73. So what does this mean? Well, this tells us that the longest valley that can be moved around the corner without tilting it is about 19.73 ft. If we look back at our answer choices, that tells us that C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Turning a corner with a pole
What is the length of the longest pole that can be carried horizontally around a corner at which a corridor that is a ft wide and a corridor that is b ft wide meet at right angles?

Key Concepts

Geometric Optimization

Pythagorean Theorem

Calculus of Variations

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