Find the area of the largest rectangle having one side on the x-axis and inscribed in a triangle formed by the lines y=x, y=0 and 3x+y=20.
If the rectangle has height x, then it has vertices
(x,0), (x,x), ((20-x)/3,x), ((20-x)/3,0)
so, its area is
a = x((20-x)/3-x) = 20x/3 - 4x^2/3
da/dx = 20/3 - 8x/3
da/dx=0 at x=5/2, and
a(5/2) = 25/3
Thank you so much!
To find the area of the largest rectangle, we need to determine its dimensions. The rectangle will have one side on the x-axis and be inscribed within the triangle formed by the lines y=x, y=0, and 3x+y=20.
Let's start by graphing the given lines and identifying the points where they intersect.
First, let's find the intersection point of y=x and y=0:
Setting the equations equal to each other, we have:
x = 0
This gives us the point (0, 0).
Next, let's find the intersection point of y=x and 3x+y=20:
Substituting y=x into the second equation, we have:
3x + x = 20
4x = 20
x = 5
Using this value of x, we can find the corresponding y-value:
y = x
y = 5
This gives us the second intersection point (5, 5).
So, the triangle formed by the lines y=x, y=0, and 3x+y=20 has vertices at (0, 0), (5, 0), and (5, 5).
Now, let's find the dimensions of the rectangle inscribed within this triangle.
The height of the rectangle will be the distance between the x-axis and the line y=x. Since the line y=x is passing through the point (5, 5), the height of the rectangle is 5.
The base of the rectangle will be the distance between the two intersection points we found previously, which is the difference between their x-coordinates. Since the intersection points are (0, 0) and (5, 5), the base of the rectangle is 5.
Therefore, the dimensions of the rectangle are 5 (base) and 5 (height).
Finally, we can calculate the area of the rectangle using the formula: Area = base * height.
Area = 5 * 5 = 25.
Hence, the area of the largest rectangle inscribed within the triangle formed by the lines y=x, y=0, and 3x+y=20 is 25 square units.
To find the largest area of a rectangle inscribed in the given triangle, we can use the concept of optimization using calculus. Here's how you can approach this problem:
Step 1: Visualize the problem
First, let's draw a graph of the lines y = x, y = 0, and 3x + y = 20. This will help us understand the boundaries of the triangle and the rectangle that we need to find.
Step 2: Represent the vertices of the rectangle
Let's consider the rectangle inscribed in the triangle. One of the sides of the rectangle will lie on the x-axis, and the opposite vertex will lie on the line y = x. We can represent the vertices of the rectangle as (0, 0), (x, 0), (0, x), and (x, x).
Step 3: Express the area of the rectangle
The area of the rectangle is given by the product of its length (x) and width (x), which is A = x * x = x^2.
Step 4: Express the constraints
We need to find the maximum area of the rectangle, given the constraints of the triangle. The constraints are as follows:
- The triangle is bound by the lines y = x, y = 0, and 3x + y = 20.
- The vertices of the rectangle lie on these lines.
Step 5: Express the area in terms of a single variable
Next, we need to express one of the variables (x or y) in terms of the other, based on the given constraints. Let's express y in terms of x using the equation 3x + y = 20: y = 20 - 3x.
Step 6: Express the area in terms of a single variable (continued)
Since one side of the rectangle lies on the x-axis, the length of the rectangle (x) should be between 0 and 20/3. Therefore, the area of the rectangle in terms of a single variable x is: A = x * (20 - 3x).
Step 7: Find the maximum area
To find the maximum area, we can take the derivative of A with respect to x, set it equal to zero, and solve for x. Let's differentiate A = x * (20 - 3x) with respect to x:
dA/dx = 20 - 6x
Setting dA/dx = 0 gives 20 - 6x = 0.
Solving this equation, we find x = 20/6 = 10/3.
Step 8: Determine the maximum area
Substitute the value of x = 10/3 back into the equation for A to find the maximum area:
A = (10/3) * (20 - 3(10/3))
= (10/3) * (20 - 10)
= (10/3) * 10
= 100/3
Therefore, the maximum area of the rectangle inscribed in the given triangle is 100/3 square units.