find the dimensions of the rectangular area of maximum area which can be laid out within a triangle of base 12 and altitude 4 if one side of the rectangle lies on the base of the triangle
thanks
The height of the rectangle can be anything from 0 to 4. Call it h.
The width (4) of the rectangle varies linearly from 12 to 0, with
w = 12 (1 - h/4)= 12 - 3h
Area = f(h) = h*w = 12h - 3h^2
dA/dx = 0 when 6h = 12
h = 2; w = 12 - 6 = 6
Amax = 12
The triangle does not have to be isosceles
To find the dimensions of the rectangular area of maximum area within a triangle, we need to understand the geometry of the problem.
First, let's draw a diagram to help visualize the situation:
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In this diagram, let's label the base of the triangle as "b" (which is 12 in this case), and the altitude as "h" (which is 4 in this case).
Now, we want to find the dimensions (length and width) of the rectangle that can be placed within the triangle in such a way that one side of the rectangle lies on the base of the triangle. Let's label the dimensions of the rectangle as "x" and "y".
To solve this problem, we'll use the concept of similar triangles. We can see that the sides of the rectangle that are parallel to the triangle's altitude (h) will also be parallel to the triangle's base (b).
Let's analyze the situation further:
- The height of the rectangle (y) is equal to the height of the triangle (h).
- The width of the rectangle (x) is equal to the length of the base of the triangle where the rectangle is placed.
Using this information, we can set up the following proportion:
y / h = x / b
Now, we substitute the known values:
y / 4 = x / 12
To find the dimensions of the rectangle with the maximum area, we need to maximize the area function. The area of a rectangle is given by the formula:
Area = length x width = x * y
We can substitute y = 4 and x = (12y) / 4 from the proportion we set up earlier:
Area = (12y/4) * y = 3y^2
Now, we have the area function in terms of a single variable, y. To find the maximum area, we can take the derivative of the area function with respect to y, set it equal to zero, and solve for y.
d(Area) / dy = 6y = 0
Solving for y gives us y = 0.
Since y = 0 does not make sense in this context (we cannot have a rectangle with zero height), we need to consider the endpoints of the possible interval. In this case, the interval is restricted to 0 ≤ y ≤ 4 (as y cannot exceed the height of the triangle).
By evaluating the area at the endpoints of the interval, we can determine which value of y gives us the maximum area.
At y = 0, the area is A = 3(0)^2 = 0.
At y = 4, the area is A = 3(4)^2 = 48.
Comparing the areas, we see that A = 48 is greater than A = 0.
Therefore, the rectangle of maximum area has dimensions:
Length (x) = (12 * 4) / 4 = 12
Width (y) = 4
So, the dimensions of the rectangular area of maximum area that can be laid out within the given triangle are 12 units (length) by 4 units (width).
That is a problem you do when refreshed, and have some time.
Here it is worked given the triangle vertexes.
You are given two point, and the altidude. For your upper vertex, x,y, choose it such that the altitude (y) is 4.
http://www.analyzemath.com/calculus/Problems/maximize_area_rectangle.html
Yours is somewhat more difficult.