Oprimization problem Right trapezoid (design of a building)

[Fibonaccy]

Hi! I tried to do this exercise but I don't know how to do it. Any ideas about how to do it or where I can learn to do that?

In the design of the building, the available surface must be used to the maximum. Target space is a plant in
trapezoidal shape, which has the following dimensions: one of the parallel walls has the dimension of 20m of
length and oblique wall is 40m. What dimensions should the plant have to make its area be
maximum?

 

[Dr.Peterson]

Hi! I tried to do this exercise but I don't know how to do it. Any ideas about how to do it or where I can learn to do that?

In the design of the building, the available surface must be used to the maximum. Target space is a plant in
trapezoidal shape, which has the following dimensions: one of the parallel walls has the dimension of 20m of
length and oblique wall is 40m. What dimensions should the plant have to make its area be
maximum?

View attachment 34704

First, it will be important for you to show us (a) what you tried, and (b) what you have learned. If you are taking a course, what is that course, and what topics have you been learning? I would expect you to learn what you need to learn in that course!

Second, the picture doesn't seem to match the problem; and the problem needs a picture in order to clarify what it means. Did you make the picture, or choose one for a similar but different problem? If we change the numbers to those in the problem, it seems like a good start.

I would expect to use calculus, though sometimes that can be avoided. Do you know any?

 

[Otis]

one of the parallel walls has the dimension of 20m of length and oblique wall is 40m

Hi Fibonaccy. The image doesn't seem to match the given dimensions. Are we supposed to change the labels 15m to 20m and 30m to 40m?

If so, then let's start with your attempt. Did you use the Pythagorean Theorem to write y in terms of x? Please show any work you've tried.
[imath]\;[/imath]

 

[Otis]

change the labels 15m to 20m and 30m to 40m

I'm not sure what happened to Fibonaccy. For other readers interested, here's my work (after changing the diagram's 15m and 30m measurements to 20m and 40m, respectively).

The building's floor plan is composed of a rectangle and right triangle. The area of the rectangular part (base×height) is 20x. The area of the triangular part (½×base×height) is ½xy. Therefore, the floor-space area can be modeled with 20x+½xy.

Before we can maximize that, we need the area expression to contain only one variable. Let's choose x. Therefore, we rewrite variable y in terms of x, using the Pythagorean Theorem.

x^2 + y^2 = 40^2

Solving for y, we get ±√(40^2 – x^2). We use the positive result because y is a distance.

y = √(1600 – x^2)

Substituting that expression for y into our area model yields:

Area = 20x + x √(1600 – x^2)/2

I'd been thinking of the property that a square is the rectangle with maximum area when the perimeter is fixed. So, I had a hunch that y=20 and worked out that x is 20√3. It turns out that is correct, but that hunch now seems invalid because the rectangle is not a square. So, I'd confirmed using a graphing calculator to zoom in on the maximum area.

The calculus method involves setting the first derivative of the area function equal to zero and solving for x, followed by using the formula for y above.

The first derivative is 20 + ½*√(1600 – x^2) – ½*x^2/√(1600-x^2)

The building's floor space is maximized when x ≈ 34.6m and y = 20m.
[imath]\;[/imath]

 

[Steven G]

Why must the 15 become 20? Why not the other parallel side be 20??

 

[Otis]

Why must the 15 become 20?

The changes were a guess on my part, corresponding to diagram labels that don't match texted dimensions. In other words, I didn't change the unknown dimensions; I'd retained those, as shown in the diagram. (You could show what happens, after swapping side lengths 20 and x.)
[imath]\;[/imath]