Hello, could someone please help me with this problem? I'm a little stuck with it. Thanks, Isaac
A rectangle is to be inscribed with its base on the x-axis and its other two vertices above the x-axis on the parabola y=9-x^2
(a) find the dimensions of the rectangle of largest area.
(b) Find the area of the largest rectangle.
Since the parabola is symmetric about the line x=0, let the rectangle have corners
(-x,0) (x,0) (x,y) (-x,y)
Since y is 9-x^2, the rectangle is
2x by (9-x^2) in width and height. The area is thus
A = 2x(9-x^2) = 18x - 2x^3
To find the value of x which gives maximum area, we want
A' = 0
A' = 18 - 6x^2
A' = 0 when x = √3
So, the rectangle is
2√3 by 6 with area 12√3
where does the 6 come from
Hi Isaac! I can definitely help you with this problem.
To solve this problem, we need to find the dimensions of the rectangle and its maximum area. Here's how we can approach each part:
(a) To find the dimensions of the rectangle with the largest area, we start by visualizing the problem. We have a rectangle with its base on the x-axis, so its height will be the distance between the x-axis and the parabola y=9-x^2.
To find the height, we need to determine the y-coordinate of the point on the parabola where the rectangle's vertices are located. Since the base is on the x-axis, the x-coordinate of both vertices will be the same. Let's call this x-coordinate "x".
Substituting "x" into the equation of the parabola, we get y=9-x^2. So the height of the rectangle is y=9-x^2.
Now, let's find the width of the rectangle. The width is simply twice the x-coordinate of the rectangle's vertices, since the base is on the x-axis. Let's call this width "w".
To maximize the area of the rectangle, we need to express the area function in terms of a single variable. The area of a rectangle is given by A = length * width. In this case, the length of the rectangle is the height, which is y=9-x^2, and the width is w.
So the area function is A = (9-x^2) * w.
Next, we can express the width "w" in terms of the x-coordinate: w = 2x.
Now we have the area function in terms of a single variable x. Let's substitute w = 2x into A = (9-x^2) * w:
A = (9-x^2) * (2x)
Expanding the expression, we have:
A = 18x - 2x^3
To find the dimensions of the rectangle with the largest area, we need to find the critical points of the area function. We can do this by taking the derivative of A with respect to x and setting it equal to zero:
dA/dx = 18 - 6x^2 = 0
Solving this equation, we get:
6x^2 = 18
x^2 = 3
x = ±√3
Since the rectangle is inscribed, we only consider the positive value of x: x = √3.
Now we have the x-coordinate of the vertices, and we can find the corresponding y-coordinate by substituting x into the equation of the parabola:
y = 9 - (√3)^2
y = 9 - 3
y = 6
Therefore, the dimensions of the rectangle of largest area are width = 2x = 2√3 and height = y = 6.
(b) To find the area of the largest rectangle, we substitute the values of width and height into the area formula:
A = (2√3)(6)
A = 12√3
So the area of the largest rectangle is 12√3.
I hope this explanation helps you understand how to solve this problem! Let me know if you have any further questions.