35. Determine the dimensions of the rectangle of largest area ...

Question

35. Determine the dimensions of the rectangle of largest area that can be inscribed in the right triangle shown in the accompanying figure.

Diagram of a right triangle with an inscribed rectangle, labeled with dimensions 5, 4, 3, w, and h.

Accepted Answer

Welcome back, everyone. Find the dimensions of the rectangle with the maximum area that can be fit inside the given right triangle as shown in the figure provided. For this problem, we're going to model the whole situation on a two-dimensional coordinate plane, assuming that point Q is the origin. We can draw a two dimensional representation of this context. So basically we're going to draw the horizontal and the vertical axis. We're going to add 3 points starting with Q, which is our origin, at 00. We can see that P is 5 units above, right? So essentially this means that P has a coordinate of X equals 0 or Y equals 5. And we can also show that R, which is 12 units to the right of the point Q, has coordinates of 12.0. So now our goal is to essentially derive the equation of the line. That connects points P and R. How do we do that? Well, essentially we need to find the slope of the line. We know that Y equals MX + B is the equation that represents the line. So our slope, which is the change in Y divided by change in X, can be found by taking the coordinates of P and R. We can take Y2S5. Subtract Y1 which is 0. And divide that by the change in X. So we're going from 0 to 12, so 0 minus 12. And this gives the C slope of -5 divided by 12. What about the Y intercept B? Well, we already know that the Y intercept is 5, so B is 5. And the equation of the line becomes Y equals -5 divided by 12 X plus 5. This is the Y coordinate and essentially it corresponds to the width of the rectangle, right? Because we have a vertex of our rectangle within that line, which is the hypotenuse of the right triangle. We also know that the length of a rectangle is going to be X, just the X coordinate itself. So now what is the area function in terms of X? We're multiplying land by width, right? So we have our land. We're multiplying that by width. We know that our length is X and our width is Y. So X multiplied by -5 divided by 12 X. Plus 5. And now we can distribute the terms. This gives us -5 divided by 12 x 2 + 5 X. Well done, so now we have the area function. We can just leave that area function for now. And we're going to optimize it. In order to optimize it for the maximum area, we have to calculate the derivative of the area function. So it's differentiate a of X using the power rule. This gives us -5 divided by 12, multiplied by 2 X because the derivative of X quad is 2 X. Plus 5 multiplied by 1, the derivative of X is 1. Therefore, the derivative is -5 divided by 6 X plus 5. So we have the derivative of the area function and we're going to set it equal to 0 to identify the critical points. Which means that 5 divided by 6 X. Plus 5 is equal to 0. Subtracting 5 from both sides, we get -5 divided by 6 X equals -5. Multiplying both sides by -1. We get 5 divided by 6 X equals 5. And then X is equal to. 5 multiplied by 6, we're multiplying both sides by 6 and dividing by 5. Which means that the critical point is x equals 6. Now that we have our critical point, we want to show that this critical point corresponds to the maximum area. So we're going to use the second derivative test. Let's identify a prime of X, which is the derivative of -5 divided by 6 X + 5. This gives us -5 divided by 6. The derivative of X is 1 plus 0, the derivative of 5 is 0 because 5 is a constant. And this is negative for any acts. And what does that mean? Well, according to the 2nd derivative test, this is a point of maximum because our function is concave down at any X, right? Well done. So for any X, including X equals 6, the second derivative is negative, so we have verified that X equals 6, the critical point itself, is the maximum point. Now that we have shown that this corresponds to. A maximum. We want to identify the dimensions of the rectangle. So we know that X, which is length, is equal to 6. And then why, which is with. Is going to be equal to. Now to calculate why, we need to rebuild the relationship between X and Y. Right? So the equation of our line was Y equals -5 divided by 12 x + 5, and we can evaluate Y is -5 divided by 12 multiplied by 6 + 5. This gives us 2.5. Well then, so we have identified the required dimensions, and we can just write those dimensions as X multiplied by Y, length multiplied by width, which is 6 by 2.5, and that will be our final answer to this problem. Thank you for watching.

35. Determine the dimensions of the rectangle of largest area that can be inscribed in the right triangle shown in the accompanying figure.

Key Concepts

Optimization

Similar Triangles

Derivative Test

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