Shortest ladder A 10-ft-tall fence runs parallel to the wall o...

Question

Shortest ladder A 10-ft-tall fence runs parallel to the wall of a house at a distance of 4 ft. Find the length of the shortest ladder that extends from the ground to the house without touching the fence. Assume the vertical wall of the house and the horizontal ground have infinite extent.

Accepted Answer

Welcome back, everyone. In this problem, a 12 ft fence is built 5 ft away from a vertical wall. Suppose the ground and the wall extend infinitely. What is the length of the shortest ladder from the ground to the wall without touching the fence? Round the answer to two decimal places. A says it's 23.33 ft. B. 8.96 ft, C, 11.67 ft, and D 17.92 ft. Now, let's try to visualize what's going on here to help us solve this problem, to help us to minimize the length of the ladder from the ground to the wall, OK? Given these conditions. So we're saying here that we have a wall. Let me just draw this here as a triangle, right? So we have a wall, OK. And here we have our ladder and then we have a fence, OK. So let me try to draw that as straight as possible here. And now we know that our fence is 5 ft away from our wall, OK? So this distance is 5 ft. And then we don't know the distance from the fence to the base of the ladder. So let's call that distance X. Let's also call the length of the ladder, OK. Let me just draw this properly here. Let's say this is our this is the length of our ladder here and let's just make some space and call it L. OK. And we also know that the world has a height, so let's call that height each. OK. And then we know that our fence is 12 ft, so let's also represent that distance on our diagram. Now, what do we know here? 12 ft. What are we trying to find here, actually? Well, if we think about it, what we really want is to find the length of the shortest ladder from the ground to the wall. So we're trying to minimize L essentially, OK? Now, what do we know about L that's related to the rest of information that we have here? Well, by the Pythagorean theorem, OK, by the Pythagorean theorem. Assuming that the base of the, the base of the wall and the ground make a right angle degree that they are perpendicular, sorry, make a right angle that they are perpendicular to each other, then by the Pythagorean theorem, L squared is going to be equal to the vertical component H2 plus the horizontal component which in this case is going to be X + 5 squad. Now to find the length of the shortest ladder, we'll need to minimize l squared, but to minimize it, it would be best to do so where we have or we have L expressed purely in terms of one variable. So let's try to express L in terms of X. To do that, we'll need to express h in terms of X. Now what do we know that can help us to do that? Well, if we look back at our diagram here, notice that we have similar triangles. We have two similar triangles or a larger triangle. With height H is similar to our smaller triangle with our height of 12 ft, OK? So using the property of similar triangles. OK, using the property of similar triangles. Then we can say that the ratio of the sizes will be the same. In other words, the ratio of the height of the larger triangle to its base X + 5 is going to be equal to the ratio of the height of the smaller triangle 12 to its base X. Which tells us then that we can rewrite H as 12 multiplied by x + 5 divided by X. I know that we have an expression for H, we can substitute it into our formula for L squad because no. This means that I. Is squared, sorry, is gonna be equal to 12. Multiplied by x + 5 divided by X squared. Plus x + 5 squared. And if we expand our first term here, if we simplify it, we could rewrite it then as 1 + 144 divided by X squared. OK. So now, this is going to be the expression for L squared. I know that we have this expression, then we can minimize L with respect to X, OK? Or minimize L2. To minimize it, that means we can differentiate L2. OK. So we're differentiating L2d with respect to X, OK? So if we differentiate it, then set our derivative to 0, we'll get a critical point and we can test if that will be a minimum value, OK? To get our minimum value of X to eventually help us solve for L. Now, in this case, OK. Notice that L squared is a product, so we can differentiate it by the product rule. Recall that the product rule of differentiation basically tells us that if we have a function Y that's equal to UV, where U and V are functions of X, then the derivative of Y with respect to X will be equal to U V plus UV where U and V are the derivatives of U and V respectively. So in our case, if we let you, OK. If we let U be equal to 1 + 144 divided by X2. And if we let V be equal to X + 5 or 2, we can differentiate both and plug them into the product to, OK? Now if we look at our expression here, the derivative of you. Is going to be equal to -288 multiplied by X to -3, OK? Likewise, the derivative of V is going to be equal to 2 multiplied by x + 5. So now in this case that means then the derivative of L squared with respect to X. Is going to be equal to U multiplied by V. Good right here. That's uh 288, multiplied by X to the negative3, OK. Multiplied by V X + 5 are squared, OK, plus. You multiplied by V. OK, so that's gonna be 1 + 144 divided by X squared. Multiplied by 2 multiplied by X + 5. Now, we can go ahead and start to simplify, OK? Now, first, uh, if, if we think about it here, we could simplify our through this term or write this as one term with a denominator of X2, OK? And if we do that, that means we can multiply through here. By 2, OK, I know when we do that, that is going to be equal to. 2 I'll write this term 1st. 2 multiplied by X + 5, multiplied by X2 + 144 divided by X2. Again, that's because we wrote X2d as a common term. We combined 1 and 144 divided by X2. OK. Next. If we write it over here, it's gonna be minus 288, multiplied by X + 5 or squared, OK, divided by X cubed. Now, we can simplify this even further and I'll leave it to you to work out the details here in the interest of time. But when we simplify here, when we break it down even further with our common denominator of X cubed, then we get the derivative of L squared with respect to X. To be equal to 2 multiplied by x + 5 multiplied by X cubed minus 720 all divided by X cubed. OK, where X is not equal to 0, which makes sense because X is a dimension, so it must be a positive value. Know that we have this, we can set our first derivative to zero. OK, setting, not setting. So setting our first derivatives to 0, OK, then. We'll get 2 multiplied by X + 5 multiplied by X cubed minus 720 all divided by X cubed to be equal to 0. And no, that means our denominate. Well, we can multiply through by Xcubed, OK. To get our numerator, the expression in our numerator to multiply, well, we can multiply through by X cubed and uh divide through by 2. OK, so let me just write that here. Multiply both sides. I just want to make sure you see what we're doing here and make it visible. So we can multiply this side by XQ divided by 2 and multiply or. Right hand side by excubed divided by 2. And when we do that, OK, then we're going to get X + 5. Multiplied by x cubed minus 720 to be equal to 0 and in that case then X is going to be equal to -5 or X is going to be equal to the q root of 720 which we could rewrite as 2 multiplied by the cube root of 90. Now, we know that X cannot be negative because it's a length, OK? So, therefore, this must mean that X is going to be equal to 2 multiplied by the Q root of 90 ft. This is a critical value, but now, we need to test if it's a minimum. We can test if it's a minimum using the first derivative test, OK? So testing if X is a minimum. We know that it will be a minimum, OK, if the derivative changes signed from negative to positive for our critical point before and after a critical point, OK? So let's choose two values, OK. First, when X is less than our critical point, uh, which we could say that let's let X equals 8. Now when X equals 8. OK, then our derivative. Is going to be equal to 2 multiplied by 8 + 5 multiplied by 8 cubed minus 720 divided by 8 cubed, which equals -169 divided by 16. OK. So this value is negative. When X is greater than 22 multiplied by the cube root of 90, OK, then we can let X be equal to 9. And if we let X be equal to 9, then our derivative, OK, again, if we substitute these values, we should find that it equals 28 divided by 81, which is positive. So notice our sign moves from negative to positive, OK. So since the sign. Changes from negative to positive. OK. Then that means. X equals 2 multiplied by the cube root of 90 is a minimum. We've confirmed that, OK? We've confirmed that this is indeed a minimum. And now that we know it's a minimum, then we can solve for L. So now, remember we said that L2d is equal to X + 52 multiplied by 1 + 144 divided by X2. So no, that means it's gonna be equal to 290 + 5 all squared multiplied by 1 + 144 divided by 290 square, OK? I know when we work that out, OK, then that means I. Will be equal to the square root of this expression, OK, if we square root both sides, and when we do that, then we get L to be approximately equal to 23.33 ft to 4 significant figures. Thus, the length of the ladder that will allow us to have a minimum or sorry, that will allow us to. The sorry, the length of the shortest ladder from the ground to the wall without touching the fence has to be 23.33 ft. Thanks a lot for watching everyone. I hope this video helped. Therefore, A, we can say A is the correct answer.

Shortest ladder A 10-ft-tall fence runs parallel to the wall of a house at a distance of 4 ft. Find the length of the shortest ladder that extends from the ground to the house without touching the fence. Assume the vertical wall of the house and the horizontal ground have infinite extent.

Key Concepts

Optimization

Related Rates

Geometric Relationships

Watch next