6. You are planning to close off a corner of the first quadran...

Question

6. You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a, 0) to (0,b). Show that the area of the triangle enclosed by the segment is largest when a = b.

Accepted Answer

Hello. In this video, we are going to be working on the following optimization problem. So we are told that in a coordinate plane, a student aims to create a triangular plot for a science project by stretching a rope 40 units long from 1.0 to 0.0.N in the first quadrant. What should be the values of M and N to maximize the area of the triangular plot? So before we approach this problem, let's just go ahead and create a picture for the given information. First, we are working on a coordinate plane, so let's go ahead and draw a coordinate plane. Next, we are told that a student is trying to create a triangular plot by stretching a rope in the first quadrant. Now, this rope is going to be connected to two points M0 and 0. N. Now, M.0 is going to be some generic point along the X axis in the positive direction. So we can label a random point to be M.0. The 0.0.N is going to be some arbitrary point along the positive Y axis. So we can label this point in some random position along the Y axis and call this 0. N. Now, the problem states that we are going to stretch a rope between these two points, and it's going to have a length of 40 units. This rope is going to create a right triangular area. Now, the length of this triangle is going to be from the origin to some point M along the X axis, so the length can be labeled as some value M. Furthermore, the height of this triangle is going to be from the origin to this arbitrary point N along the Y axis. So this triangle is going to have some arbitrary height N. Now, what is an equation that we can use in order to try and define the values of M and N that maximize the area? Well, one equation that we have. is going to be the area of the triangle, and the area of the triangle is given to us as the area is equal to one half, multiplied by the base, multiplied by the height. Since our base is going to be M and our height is going to be N, if we plug in these values for the area, we get area is equal to 12 multiplied by M, multiplied by N. Now, in order to find a maximum value, we want to go ahead and find a critical point of this function. A is equal to 1/2 M and N. But before we do that, let's go ahead and try to rewrite the function in terms of either M or N. Now, we know, we are going, well, first we are going to create a constraint for this, for this problem. We know that the distance between M and N is going to be 40, but the arbitrary distance formula is going to be the square root of M2 plus N2, and we know that the output of this distance is going to be 40. So we can go ahead and use this distance formula to create a constraint of N in terms of M. In order to do this, we are going to solve for N in this distance formula. We will start by squaring both sides of this equation, leaving us with M2 plus N2 is equal to 1600. If we subtract M2 to both sides of this equation, we get N2 equal to 1600 minus M2. And if we take the square root of both sides of this constraint formula, we get N is equal to the square root of 1600 minus M2. What we've created here is a constraint formula for N, and if we substitute this into our area formula, we can rewrite the area to be area is equal to 1/2 multiplied by M, multiplied by the square root of 1600 minus M squared. Now, why is this useful to us? The reason why this is useful is because now if we go through this equation and take the derivative with respect to M, this is going to give us a derivative that we can use to find a critical point for M, and we can show that this critical point is going to be a value of M that maximizes the area. So let's go ahead and take the derivative of our area function with respect to M. On the left hand side, the derivative of the area is going to be D A DM. And on the right hand side, by using the quote, the power rule, we can go ahead and write the derivative of the right hand side to be 800 minus M2d divided by the square root. Of 1600 minus m squared. Now, in order to find a critical point for M, we want to find the value of M that causes this numerator to become zero. So, in order to do this, we are going to take our numerator, 800 minus M2, set it equal to 0, and solve for M. We will go ahead and add M to both sides of this equation, leaving us with M2 equal to 800. And by taking the square root of both sides of this equation, we get M equal to 20 multiplied by the square root of 2. Now we need to verify that this value of M is going to be a value that is a maximum. Now, 20 multiplied by the square root of 2 has an approximate value of 28.3. If we plot this on a number line. And plot 28.3 or 20 square root of 2, we can go ahead and pick values to plug into the derivative and prove that this is going to be a maximum. A critical point or a value that we can pick to plug into the derivative is going to be 29, and another point will be 27. If we plug in 29 into the derivative, A of 29 will give us the value of 2.4. Oh, apologies, not 2.4. It will give us the value of negative 1.5. So we know that the derivative will be decreasing to the right of the critical point. And if we plug in the value of 27 into the derivative, we get the value of 2.4, which is going to be positive. So, the derivative is going to shift from increasing to decreasing as we pass through this critical point, which shows that M is going to be a maximum value that maximizes the area. Now that we know that M equal to 20 square root of 2 is a maximum, we can take this value of M and plug it back into our original constraint, and that is going to allow us to solve our N that gives us a maximum area. So we will go ahead and use N equal to the square root of 1600 minus M2d. So we have N is equal to the square root of 1600 minus 20 multiplied by the square root of 2 squared. 20 multiplied by the square root of 2 squared will simplify to be 800, so we have N is equal to the square root of 1600 minus 800, which is going to leave us with the square root of 800. And the square root of 800 will simplify to be 20 multiplied by the square root of 2, and this is going to be a maximum value for N. So what does this mean? Well, what this means is that in order to find the maxim or in order to maximize the area of the triangle, the value of M and N must have the same value, and that value is going to be 20 square root of 2. And this is going to be the solution to the problem. So I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

6. You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a, 0) to (0,b). Show that the area of the triangle enclosed by the segment is largest when a = b.

Key Concepts

Equation of a Line

Area of a Triangle

Optimization

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