The ladder problem What is the approximate le...

Question

The ladder problem What is the approximate length (in feet) of the longest ladder you can carry horizontally around the corner of the corridor shown here? Round your answer down to the nearest foot.

Accepted Answer

Hello there. Today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A painter needs to carry a long wooden plank horizontally through an L-shaped hallway. The hallway consists of a 5-foot wide passage leading into a 9-foot wide section. What is the approximate length in units of feet of the longest plank that can be maneuvered around the corner without tilting? Round your final answer to two decimal places. Fantastic. So it appears for this particular prong, we're trying to figure out the approximate length in units of feet. Of the longest plank that can be maneuvered around this particular L-shaped hallway's corner without tilting. So we're trying to figure out the approximate length of this longest plank that can be maneuvered around this corner without tilting, and that's what our final answer we're trying to solve for is what is this length value? And then we're rounding our final answer to two decimal places. So now that we know what we're solving for, let's read off our multiple choice answers to see what our final answer might be, noting that they're all in the same units of feet. A is 19.52, B is 17.25, C is 10.30, and D is 23.45. Fantastic. So first off, let us create a diagram to help us better visualize this problem so we can see what exactly is happening. So with that said, when we produce a diagram of this particular problem to help us visualize what's happening, we could see, and let's call L to be the length of the plank. We have one length compared, so right now we have our our this L shaped hallway. And the hallway consists of a 5 ft wide passage that leads into a 9 ft wide section. So as you can see, it forms almost like a rectangular shape when we look at our dotted lines, and then where the planks are diagonal, which once again, this is the length of the plank, which is going to be represented in blue, it will form an angle theta with this dotted line that forms a triangle that has 5 ft. On going up for the length and then 9 ft going toward the width. Awesome. And then we have another length for when it's a more narrow part of the hallway, which is L1, which also forms an angle theta with the ground. So as you can see, we have this L-shaped hallway that produces two different triangles, and we're gonna be having to recall and use trigontric identities to help us solve for this particular problem. So, with that said, based on our figure, we can go ahead and conclude that L1. Of theta is going to be equal to 5 divided by sin of theta. Fantastic. We can also go ahead and write that L2 of theta is going to be equal to 9 divided by cosine of theta, and once again we could write this based on our trigonometric identities. So now we need to note that the total length of the plank L can be represented by the following equation that states that L of theta. is going to be equal to L1 of theta plus L2 of theta, which is going to be all equal to, so L of theta is equal to 5 divided by sine of theta plus 9 divided by cosine of theta. Fantastic. That's when we plug in our values for L1 theta and L2 theta into our expression. Fantastic. So now in order to determine the longest plank that could be maneuvered around the corner without tilting, we need to minimize our L of theta value. So as we should recall a note, the minimum value corresponds to the longest plank that can still be fitted through the corner without exceeding the corner's constraints. So, in order to minimize Elaheta, as we should recall, we need to differentiate LA theta with respect to theta. So we need to recall and apply the quotient rule of differentiation, which as we should. Call states that D by D theta of U divided by V is going to be equal to Up or multiplied by U. So V multiplied by u minus U multiplied by V all divided by V squared. So, with that said, when we perform the differentiation, we can write that theta or this prime symbol denotes the first derivative of Lla theta, and this will be equal to sin of theta. Multiplied by 0, minus 5 multiplied by cosine of theta. All divided by sin squared of theta. Plus, no sign of theta. Multiplied by 0, or multiplied by 0, minus. 9 or rather -9 multiplied by negative sign of theta. All divided by Cosine squared. Of theta. And now when we simplify this expression down to its most simple form using a little bit of algebra, we will find that L of theta is going to be equal to -5 multiplied by cosine to the power of 3 of theta plus 9 multiplied by sin to the power of 3 theta. All divided by drum roll. Sine squared of theta multiplied by cosine squared of theta. So once again, I said we used a little bit of algebra, but once we perform the differentiation and we use a little bit of algebra to rearrange things to write it in write our L theta in its most simple form, we will find that it's equal to -5 multiplied by cosine to the 3 of theta plus 9 multiplied by sin to the power 3 theta, all divided by sin square to theta multiplied by cosine square to theta. So now, We need to set L of theta equal to 0. So the derivative will be 0 when the numerator is 0. So therefore we can go ahead and write that negative 5 multiplied by cosine to the power of 3 of theta plus 9 multiplied by sin to the power 3 of theta is equal to 0. So, what we need to do now is we're ultimately trying to take this expression that we just wrote, which is the negative 5 multiplied by cosine to the 3 of theta plus 9 multiplied by sin to the power of 3 of theta equals 0, and we're going to be rearranging it to isolate and solve for theta all by itself. And when we do that, we will find that theta is going to be equal to inverse tangent of the cube root of 5 divided by 9, which is going to be approximately equal to 0.68805, and this is when it's rounded to 12345 decimal places, when it's rounded to 5 decimal places. So now our next step is we need to confirm whether theta equals inverse tangent of the cube root of 5 divided by 9 corresponds to a minimum using the first derivative test. So we need to note that when theta is less than Inverse tangent of the cube root of 5 divided by 9. We need to let theta equal 0.60. So, when we plug in a value of 0.60 in for theta, we will find that L Of 0.60 is going to be equal to -5 multiplied by cosine to the power 3 of 0.6 plus 9 multiplied by sin to the power of 3 of 0.60. And then we need to divide this all by Sine squared of 0.60 multiplied by cosine squared of 0.60. Fantastic. And when we plug in that big old long expression into our calculator, we will find that it's approximately equal to -5.48328, and once again, that's when we round to 5 decimal places. So now, similarly, When we do, for when tangent, I mean, sorry, when I say that theta, we're gonna say that theta is greater than inverse tangent of the cube root of 5 divided by 9, and we need to let Theta equals 0.70. So instead of rewriting our expression not plugging in the variable, I'm just going to give you the answer. So L of 0.70 is going to be approximately equal to 0.69672, once again rounding the five decimal places when you plug in a value of 0.70 indoor expression instead of 0.60. So now we need to note that since the sign of the derivative changes from negative to positive, that will mean that theta equals inverse tangent of the cube root of 5 divided by 9 corresponds to a minimum value. So now we need to solve for our value of L. So when we do that, we will find that L of inverse tangent of the cube root of 5 divided by 9. It's going to be equal to 5 divided by sign of. Inverse tangent of the cube root of, so once again, this is the cube root. Of 5 divided by 9. Fantastic. So now we need to add 9 divided by cosine of inverse tangent of the cube root of 5 divided by 9, fantastic. So, with that said, when we perform that calculation, We will find That L. So this value, so our final answer is going to be approximately equal to when we round to two decile places like our final answer asks us to, it's going to be approximately equal to 19.52 ft, and that is our final answer. Hooray, we did it. So therefore, the approximate length in units of feet of the longest plank that can be maneuvered around the corner without tilting is 19.52 ft. And we also need to note that if L is represented as, and let's make a note of this up in red up above. So let us note that L of theta is going to be equal to 5 divided by cosine of theta. So 5 divided by cosine of theta plus 9 divided by sine of theta. Fantastic. So let us note that if L is represented as L of theta is equal to 5 divided by cosine theta plus 9 divided by sin theta, the answer will be the same. So with that said, let's look at our multiple choice answers. The correct answer, looking at our multiple choice answers has to be the letter A, which once again is 19.52 ft. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

The ladder problem What is the approximate length (in feet) of the longest ladder you can carry horizontally around the corner of the corridor shown here? Round your answer down to the nearest foot.

Key Concepts

Optimization

Geometric Relationships

Calculus of Functions

Watch next